Troubleshooting - A Guide to MATLAB for Beginners and Experienced Users - page 271

They say a brave man dies but a single time, but a coward dies a
thousand deaths. But the person who said that probably never
encountered a Cardassian. Long live Picard!

Problem 2

(a)

Consider the status of the account on the last day of each month. At the
end of the first month, the account has M + M × J = M (1 + J ) dollars.
Then at the end of the second month the account contains
[ M (1 + J )](1 + J ) = M (1 + J ) 2 dollars. Similarly, at the end of n months,
the account will hold M (1 + J ) n dollars. Therefore, our formula is

T = M (1 + J ) n .

(b)

Now we take M = 0 and S dollars deposited monthly. At the end of the
first month the account has S + S × J = S (1 + J ) dollars. S dollars are
added to that sum the next day, and then at the end of the second month
the account contains [ S (1 + J ) + S ](1 + J ) = S [(1 + J ) 2

+ (1 + J )]
dollars. Similarly, at the end of n months, the account will hold

S [(1 + J ) n +···+ (1 + J )]

dollars. We recognize the geometric series — with the constant term “1”
missing, so the amount T in the account after n months will equal

T = S [((1 + J ) n + 1

− 1) / ((1 + J ) − 1) − 1] = S [((1 + J ) n + 1 − 1) / J − 1] .

(c)

By combining the two models it is clear that in an account with an initial
balance M and monthly deposits S , the amount of money T after n
months is given by

T = M (1 + J ) n

+ S [((1 + J ) n + 1

− 1) / J − 1] .

(d)

We are asked to solve the equation

(1 + J ) n

= 2

withthe values J = 0 . 05 / 12 and J = 0 . 1 / 12.