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Compound Interest Help

By — McGraw-Hill Professional
Updated on Aug 31, 2011

Introduction Compound Interest

Yet a third illustration of exponential growth is in the compounding of interest. If principal P is put in the bank at p percent simple interest per year then after one year the account has

Transcendental Functions 6.5 Exponential Growth and Decay

dollars. [Here we assume, of course, that all interest is reinvested in the account.] But if the interest is compounded n times during the year then the year is divided into n equal pieces and at each time interval of length 1/ n an interest payment of percent p/n is added to the account. Each time this fraction of the interest is added to the account, the money in the account is multiplied by

Transcendental Functions 6.5 Exponential Growth and Decay

Since this is done n times during the year, the result at the end of the year is that the account holds

Transcendental Functions 6.5 Exponential Growth and Decay

dollars at the end of the year. Similarly, at the end of t years, the money accumulated will be

Transcendental Functions 6.5 Exponential Growth and Decay

Let us set

Transcendental Functions 6.5 Exponential Growth and Decay

and rewrite (*) as

Transcendental Functions 6.5 Exponential Growth and Decay

It is useful to know the behavior of the account if the number of times the interest is compounded per year becomes arbitrarily large (this is called continuous compounding of interest ). Continuous compounding corresponds to calculating the limit of the last formula as k (or, equivalently, n ), tends to infinity.

We know from the discussion in Subsection 6.2.3 that the expression (1 + 1/ k ) k tends to e . Therefore the size of the account after one year of continuous compounding of interest is

P · e p /100 .

After t years of continuous compounding of interest the total money is

Transcendental Functions 6.5 Exponential Growth and Decay

Examples

Example 1

If $6000 is placed in a savings account with 5% annual interest compounded continuously, then how large is the account after four and one half years?

Solution 1

If M ( t ) is the amount of money in the account at time t , then the preceding discussion guarantees that

M ( t ) = 6000 · e 5 t /100 .

After four and one half years the size of the account is therefore

M (9/2) = 6000 · e 5·(9/2)/100 ≈ $7513.94.

Example 2

A wealthy woman wishes to set up an endowment for her nephew. She wants the endowment to pay the young man $100,000 in cash on the day of his twenty-first birthday. The endowment is set up on the day of the nephew’s birth and is locked in at 11% interest compounded continuously. How much principal should be put into the account to yield the desired payoff?

Solution 2

Let P be the initial principal deposited in the account on the day of the nephew’s birth. Using our compound interest equation (**), we have

100000 = P · e 11·21/100 ,

expressing the fact that after 21 years at 11% interest compounded continuously we want the value of the account to be $100,000.

Solving for P gives

P = 100000 · e −0.11·21 = 100000 · e −2.31 = 9926.13.

The aunt needs to endow the fund with an initial $9926.13.

You Try It : Suppose that we want a certain endowment to pay $50,000 in cash ten years from now. The endowment will be set up today with $5,000 principal and locked in at a fixed interest rate. What interest rate (compounded continuously) is needed to guarantee the desired payoff?

Find practice problems and solutions for these concepts at: Transcendental Functions Practice Test.

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