# Present Value    
# 
# In the continuous compound interest formula,  
# A = P*e^(rt),
# P is referred to as the 'present value' of A t years from
# now if money grows at  100r percent yearly with continuous
# compounding of interest.  So for example, the 
# present value of 50,000 dollars to be paid in 5 years if 
# money grows at 6 percent is P=50,000*e^{-.06*5)=37,040 dollars.
# 
# Suppose we are paid 50,000 dollars a year for 20 years
# from an account starting with P dollars and paying
# r percent interest compounded continously.  How large
# should P be at least?  That is, what is the present value
# of the money we will receive?
# 
# Solution: Assume we get the intitial 50,000 dollars
# right off the bat.  So the present value is
# the sum of the present values we need  to make the
# payments, which is   
# 
# 50000 + 50000*exp(-.06) +  ... + 50000*exp(-0.06*19) 
#  
# We can sum this easily in Maple.
# 
# 
> Sum(50000*exp(-.06*i),i=0..19); evalf(");

                              19
                            -----
                             \
                              )   50000 exp( - .06 i)
                             /
                            -----
                            i = 0

                                   599982.9924
# 
# If the payments were dribbled out over the year (with 
# say n payments) so that
# the total amount paid out each year was 50000, then the 
# present value would be       
> pv := n ->Sum(50000/n*exp(-.06/n*i),i=0..19*n);  

                                 19 n
                                -----
                                 \          exp( - .06 i/n)
                     pv := n ->   )   50000 ---------------
                                 /                 n
                                -----
                                i = 0
# So for example  using monthly payments
> pv(12); evalf(");

                       228
                      -----
                       \
                        )   12500/3 exp( - .005000000000 i)
                       /
                      -----
                      i = 0

                                   569568.2857
# pv(n) is a Riemann sum for 
> 
> Int(50000*exp(-.06*t),t=0..19);evalf(");

                             19
                             /
                            |
                            |   50000 exp( - .06 t) dt
                            |
                           /
                           0

                                   566817.4818
# and so the integral (which is easy to evaluate by the
# fundamental theorem)  can be used to approximate the
# sum (which is messy to evaluate by hand).
# 
# So if I want to estimate how much I need to have in
# a retirement account drawing a conservative .06 per
# cent interest so that I can take out 4000 a month
# for 30 years, the least I need is approximately
> Int(12*4000*exp(-.06*t),t=0..30);evalf(");

                             30
                             /
                            |
                            |   48000 exp( - .06 t) dt
                            |
                           /
                           0

                                   667760.8894
# Whereas the exact least amount I need is.
> Sum(4000*exp(-.06/12*i),i=1..30*12); evalf(");

                        360
                       -----
                        \
                         )   4000 exp( - .005000000000 i)
                        /
                       -----
                       i = 1

                                   666092.8784
#