24 the Exponential and Natural Logarithm Functions

24 The Exponential and Natural Logarithm Functions.

In this section we look at a very commonly used exponential and logarithm functions which are the inverse of each other and are denoted as ex and ln(x). By the end of this section, you should have the following skills:

• An understanding of the deﬁnition of the exponential and logarithm functions and their properties.

• An understanding of use of the exponential and logarithm functions in ﬁnance, including compound interest.

• Calculate an approximation of return on interest using the compound interest formula.

• An understanding of the properties of the exponential function and the constant e.

• Sketch the graphs y = ex, y = e−x and y = ln(x).

24.1 The exp(x) and ln(x) Functions In this section we examine two very important functions, the exponential function exp(x) and the natural logarithm function written as either of

ln(x) = loge(x)

The following examples show that both of these functions are important in ﬁnance. They also occur in many other scientiﬁc subjects and engineering. Note that exp(x) and ln(x) are inverse functions in that exp(ln(x)) = x, for x > 0 and ln(exp(x)) = x for all x.

1 24.2 Compound Interest In this section we consider the return earned by investing money under compound interest. We show how to obtain a formula for the return and then describe how the exponential function can be used to approximate the return.

24.2.1 Return on Compound Interest and the Compound Interest Formula You invest £p which over a year earns interest at r% per annum. At the end of the year we have: r r p + p = p 1 + . 100 100 r Now suppose the rate is quoted instead as a compounded half-year rate 2 %. Applying this rate twice in the year we get r/2 p 1 + 100 at the end of the ﬁrst half and then at the end of the year r/2 r/2 r/2 r/22 p 1 + + p 1 + = p 1 + . 100 100 100 100 r Suppose now the rate is quoted at 3 % compounded every 4 months. Then this gives r/33 p 1 + . 100

So if the interest rate is r/n% compounded n times in a year then we obtain r/nn p 1 + 100 at the end of the year. This is the Compound Interest Formula.

24.2.2 Approximating the Compound Interest Formula Increasing n to a large value (for example, quoting the interest rate every second) then the return is approximated by the limit r/nn p 1 + 100

2 as n → ∞. We shall give some examples to show that this can be a good approximation and now we show that the limit is a well known function.

24.3 The Exponential Function We see that calculating the return from compounded interest rates has lead to an expression of the form r n 1 + n where r is the annual interest rate. If n is large then we can approximate this by the limit as n → ∞. This is given a special notation: xn exp(x) = lim 1 + n→∞ n and is called the exponential function and we describe this in more detail below, showing that exp(x) = ex for a number e. Your calculators have this function (sometimes as the inverse of the natural logarithm ln(x)). You can use exp(x) to approximate compound interest as follows: Let r% be the rate quoted every d days, then the amount A invested over n days will result in a return of r n A 1 + . 100 This can be approximated by rn A exp 100 as rn rn/100n r n A exp ≈ A 1 + = A 1 + . 100 n 100

Example 1 An interest rate is set daily at 0.015%.

(a) Calculate the interest you receive on £100, 000 invested over one year (365 days).

3 (b) How does this compare with the interest gained from a simple interest rate of 5.475% = 365 × 0.015% over one year?

Solution.

(a) The return is

0.015365 100, 000 1 + = 105, 627.22 100

so the interest received is 5, 627.22.

(b) The yearly rate 5.475 gives a return of 105, 475 so the interest obtained is 5, 475.

0.015 × 365 100, 000 × exp = 105, 627.65. 100

Hence this approximation to the return only overestimates the true value by 0.43 pounds.

Example 2 An interest rate is set daily at 0.0175%.

(a) Calculate the interest received on £1, 000, 000 over half a year (182 days).

(b) How does this compare with the interest gained from a simple rate of 3.185% = 182 × 0.0175% over the half year (with 182 days giving a half-year)?

4 (c) Approximate the daily rate interest over the half year using exp(x) and comment on the error.

Solution.

(a) The return (to 2 decimal places) is

0.0175182 1, 000, 000 1 + = 1032359.76 100

so the interest received is 32, 359.76.

(b) The half-yearly rate 3.185% gives an interest of 31, 850.

0.0175 × 182 1, 000, 000 × exp = 1032362.64 100

giving an approximate interest of 32,362.64. Hence this approximation to the interest overestimates the true value by £2.88.

Exercise 1 An interest rate is set daily at 0.02%.

(a) Calculate the interest received on £250, 000 over one year (365 days).

(b) How does this compare with a simple interest rate of 7.3% = 365 × 0.02% over one year?

5 Solutions to exercise 1

(a) The return is

0.02365 250, 000 1 + = 268, 930.67 100

so the interest received is 18, 930.67.

(b) The yearly rate 7.3% gives a return of 1.073 ∗ 250, 000 = 268, 250 so the interest obtained is 18, 250.

0.02 × 365 250, 000 × exp = 268, 932.63. 100

Hence this approximation to the return overestimates the true value by 1.96 pounds.

Exercise 2 An interest rate is set daily at 0.0135%.

(a) Calculate the interest received on £1, 500, 000 over half a year (182 days).

(b) How does this compare with a simple rate of 2.457% = 182×0.0135% over the half year?

6 and comment on the error.

Solutions to exercise 2

(a) The return is

0.0135182 1, 500, 000 1 + = 1, 537, 308.95 100

so the interest received is 37, 308.95.

(b) The half-yearly simple interest rate 2.457% gives a return of 1.02457∗ 1500000 = 1, 536, 855 and hence an interest of 36, 855.

0.0135 × 182 1, 500, 000 × exp = 1, 537, 311.49 100

and so an approximate interest of 37, 311.49. Hence this approximation to the return overestimates the true value by £2.54.

24.4 The constant e In this section we show that exp(x) is an exponential function i.e. it is of the form exp(x) = ax, and ﬁnd the constant a. In order to get more information on this function we let n = xm in the formula xn exp(x) = lim 1 + n→∞ n

7 and we get: xn exp(x) = lim 1 + n→∞ n x xm = lim 1 + m→∞ xm 1 xm = lim 1 + m→∞ m 1 mx = lim 1 + m→∞ m Write 1 n e = exp(1) = lim 1 + . n→∞ n e ≈ 2.718281828 to 9 decimal places which we can check with our calculators. Hence we have that exp(x) = ex and is an exponential function.

24.5 Properties of the Exponential Function y = exp(x) has domain all x and has range y > 0. As exp(x) = ex we can use the laws of indices to get: xn • exp(x) = ex = lim 1 + . n→∞ n • exp(x + y) = exp(x) exp(y) as ex+y = exey. • exp(−x) = 1/ exp(x). • exp(x − y) = exp(x)/ exp(y). • (exp(x))y = exp(xy). • exp(0) = 1. • exp(x) > 0 for all x. • exp(x) ≤ 1 for all x ≤ 0. • exp(x) → ∞ as x → ∞. • exp(x) → 0 as x → −∞.

8 Graph of the exponential function y = ex.

The following graph is y = exp(−x) = e−x = 1/ex = 1/ exp(x).

Graph of the exponential function y = e−x.

9 24.6 The Natural Logarithm Function

x If y = exp(x) = e then x = loge(y) = ln(y) This is the natural logarithm function with base e.

Graph of the natural logarithm y = ln(x).

24.7 Properties of the Natural Logarithm Function The natural logarithm function y = ln(x):

1. has domain all x such that x > 0, and the range is all y.

2. is the inverse function of exp(x) i.e. exp(ln(x)) = x for all x > 0 and

3. ln(exp(x)) = x for all x.

4. satisﬁes all the usual properties of logarithm functions as follows, where a > 0 and b > 0:

ln(ab) = ln(a) + ln(b). ln(a/b) = ln(a) − ln(b). ln(ac) = c ln(a).

10 Example 3 Suppose that £5, 000 is invested, how long in months before the investment increases to £6, 000 if it earns an annual rate of interest 2.4% compounded monthly? Solution.

Since the period is monthly we have to work out the interest rate per payment period and this is 2.4/12 = 0.2% By the compound interest formula the original investment of £5, 000 realises an amount 0.2 n A = 5000 1 + 100 after n months. We want the value of n such that

0.2 n 5000 1 + = 6000. 100

So 0.2 n 5000 1 + = 6000 ⇒ 100 0.2 n 1 + = 6000/5000 = 1.2 ⇒ 100 n ln(1.002) = ln(1.2) on taking natural logarithms of both sides ⇒ ln(1.2) n = = 91.25 to 2 decimal places. ln(1.002)

On rounding up to the next whole number this gives 92 months.

Exercise 3 Suppose that £50, 000 is invested, how long in months before the investment increases to £60, 000 if it earns an annual rate of interest 2.28% compounded monthly?

11 Solutions to exercise 3 Since the period is monthly we have to work out the interest rate per payment period and this is 2.28/12 = 0.19%. The original investment of £50, 000 realises an amount

0.19n A = 50, 000 1 + 100 after n months. We want the value of n such that 0.19n 50000 1 + = 60000. 100

So 0.19n 50000 1 + = 60000 ⇒ 100 0.19n 1 + = 60000/50000 = 1.2 ⇒ 100 n ln(1.0019) = ln(1.2) on taking natural logarithms of both sides ⇒ ln(1.2) n = = 96.05 to 2 decimal places. ln(1.0019)

On rounding up to the next whole number this gives 97 months.

Exercise 4

(a) Show that if x > 0 then exp(x ln(x)) = xx.

(b) Show that if x > 0 then exp(x − 2 ln(x)) = exp(x)/x2.

12 (c) Find the least integer n such that

0.5n e0.5 − 1 + < 0.001. n

Solutions to exercise 4

(a) exp(x ln(x)) = (exp(ln(x)))x = xx. Note that x > 0 as otherwise ln(x) is not deﬁned. Similarly for the next question.

(b) exp(x − 2 ln(x)) = exp(x) exp(− ln(x2)) = exp(x)/ exp(ln(x2)) = exp(x)/x2.

So we compute the left hand side for n = 1, 2,... and stop the ﬁrst time that the left hand side is greater than 1.638721.

(i) n = 1 gives (1 + 0.5)1 = 1.5. (ii) n = 2 gives 0.52 1 + = 1.5625. 2

13 (iii) n = 3 gives 0.53 1 + = 1.587963. 3 ......

(iv) n = 20 gives 0.520 1 + = 1.638616. 20 (v) n = 21 gives 0.521 1 + = 1.639089. 21 Hence n = 21 is the least integer such that

0.5n e0.5 − 1 + < 0.001. n