Answers1 - EC3070 FINANCIAL DERIVATIVES Exercise 1 1 A...

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EC3070 FINANCIAL DERIVATIVES Exercise 1 1. A credit card company charges an annual interest rate of 15%, which is effective only if the interest on the outstanding debts is paid in monthly instalments. Otherwise, the interest charges are compounded with the borrowings. What would be the effective annual rate of interest if £ Q are borrowed at the beginning of the year and repaid with interest at the end of the year? Answer. Interest is charged each month on the outstanding loan at the rate of 15 / 12 = 1 . 25%. Therefore, after one year, the total amount that must be repaid is $ Q (1 + 0 . 0125) 12 = $ Q 1 . 161 , which gives an effective annual rate of interest of 16%. Using log tables, I compute (1 + 0 . 0125) 12 as Antilog { 12 × log(1 . 0125) } = Antilog { 12 × 0 . 0054 } = Antilog { 0 . 0640 } = 0 . 0159 , which gives 16% approximately. (In fact, for this calculation, four-figure base-10 logarithms are insufficiently accurate.) The alternative is to enter 1.0125 in your calculator to do the multiplication twelve times over. 1
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EC3070 Exercise 1 2. How long will it take to double your investment if you receive an annual rate of interest of 5% and if the interest is compounded with the principal? Answer. The equation to be solved is (1 + r ) n = 2 where r = 0 . 05 is the rate of interest. The solution, using log tables, is n = log2 log(1 + r ) = log2 log(1 . 05) = 0 . 3010 0 . 0212 = 14 . 2 , which is confirmed by my computer. My calculator tells me that 1 . 05 15 = 2 . 0789 2
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The Natural Number The natural number e. The number e = { 2 . 7183 . . . } is defined by e = lim( n → ∞ ) 1 + 1 n n . The binomial expansion indicates that ( a + b ) n = a n + na n 1 b + n ( n 1) 2! a n 2 b 2 + n ( n 1)( n 2) 3! a n 3 b 2 + · · · . Using this, we get 1 + 1 n n = 1 + n 1 n + n ( n 1) 2! 1 n 2 + n ( n 1)( n 2) 3! 1 n 3 + · · · . Taking limits as n → ∞ of each term of the expansion gives lim( n → ∞ ) 1 + 1 n n = 1 0! + 1 1! + 1 2! + 1 3! + · · · = e.
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The Natural Number The expansion of e x . There is also e x = lim( p → ∞ ) 1 + 1 p px = lim( n → ∞ ) 1 + x n n ; n = px. Using the binomial expansion in the same way as before, it can be shown that e x = x 0 0!
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