A sequence is a set, indexed by a natural number, traditionally either i or n. A progression is a finite
sequence. A series is a sequence of partial sums of a progression.
For conformity with the convention in the C language of computer programming, we take i to be in integer in the closed interval [0, n], where n is
any given (fixed) non-negative integer.
We define the arithmetic progression recursively as
and the arithmetic series as
where, by the definition of series as the partial sums, A(0) = 0.
Then. by Mathematical Induction, we have the arithmetic progression as the sequence
and the arithmetic series as the sequence
For convenience, we may write it in either of these two additional forms
Any of these equations may be solved algebraically (in closed form) for each of its variables.
We define the geometric progression recursively as
and the geometric series as
where, by the definition of a series as partial sums, G(0) = 0.
Then by Mathematical Induction, we have the geometric progression as the sequence
and the geometric series as the sequence
For convenience, we may write the numerator in either of these two additional forms
Any of these equations may be solved for each of its variables. Except that the solution of the series for r requires, for instance, the
Newton's method.
The limit of the geometric series, as i increases without bound, is
provided that r is within the radius of convergence, one.
In these interest formulae, P is the principal, (r - 1) is the interest for a compounding interval, and n is the amount of compounding intervals.
No one will lend you money at simple interest; but it is conventional for a textbook to give this formula.
The interest (r - 1) scales with the length of the compounding interval. Often this scaling is miss-applied to compound interest, for the benefit of the lender. One may solve algebraically for each of the variables.
The compound interest formula is
(Follow these hyper-links for the exponential and the logarithmic functions.) Write the compound interest as
and expand by the binomial theorem to yield
Observe that the first two terms are the simple interest.
Consider
From the limit of an exponential, the limit of Cm, as m approaches zero, is
Call this r the geometric ratio at zero time ro and equate Co to C. Then the ro is
The ln(r) = ro - 1 scales with the length of the compounding interval. One may solve (in closed form, in terms of the exponential and logarithmic functions) for each of the variables.
We define the annuity sequence recursively as
where, by convention, we take A(0) = 0. The lower-case p is called the payment.
Then either by observing that this is a geometric series, or else by Mathematical Induction, we have the annuity sequence
The present value of the annuity is obtained by discounting the annuity by the compound interest
The limit of the present-value of the annuity, as i increases without bound, is
provided that r is within the radius of convergence, one. Since the interest-rate is equal to (r - 1), it has to be strictly positive.
The earned future-value of the annuity is obtained by compounding the annuity over the remaining intervals
Newton's method may be employed to solve for the interest r - 1. One may solve (in closed form, in terms of the exponential and logarithmic functions) for each of the other variables. For the Newton's method, multiply by (r - 1) and transpose everything to one side of the equals sign. Then we have:
| function y = 0 | its derivative y' | |
|---|---|---|
| Annuity | A (r - 1) - p (r^i - 1) | A - p i r^(i - 1) |
| Present value | PV (r - 1) - p (1 - r^(- i)) | PV - p i r^(-i -1) |
| Future value | FV (r - 1) - r^n p (1 - r^(- i)) | FV - p n r^(n - 1) + p (n - i) r^(n - i - 1) |
Then the next iterative value of x is x = xo - yo / y'o for the linear Newton's method. Observe that the future-value for n = 0 is the present value..
A payment which occurs in the future (or past -- then the time is negative), is said to be discounted to its present value. If it is a single payment, one employs the present value of a compound interest. If it is a set of periodic payments, one employs the present value of an annuity. However, in each case, the interest rate for discounting is the time-value-of-money, ordinarily taken as the effective interest on the thirty-year U. S. Treasury bonds.. If the extrapolated zoroth payment of the annuity does not occur now, then the present value of the annuity has to be discounted further, employing the present value of a compound interest. In this case, the time might be negative.
Payments are capitalized as follows: Any payments which occur now are taken at face value. Any future (or past) payments, whether single or periodic, are discounted, as described previously. The IRS does not permit the capitalization of the value of the land of a real-estate purchase. The easiest way to perform this computation is to capitalize the down-payment and all of the periodic payments, for both the land and the improvements. Then subtract the value of the land.
A set of loan payments constitutes an annuity. Compute it as the present value of an annuity. Thus, the loan-payments p and present-value pv for a principal P, at an interest rate b ,are given by
p = P (r - 1) / (1 - r^(-i)) = P b / (1 - (1 + b)^(-i))
pv = (P b / (1 - (1 + b)^(-i))) ((1 - (1 + d)^(-i)) / d)
limit, as j increases without bound, of the pv = P b / d,
where the convergence criteria are that 0 < interest-rate and 0 < d. Please observe that this limit is displayed for
comparison with the other limits. The actual loan-payments hopefully would terminate.
Since, this interest-rate ordinarily would be higher than the discounting-rate, we have to discount the payments, as shown here but derived in the
next section.
Let us derive the formula for the present-value pv of an annuity directly. It is the geometric series of the payments p -- for example, the repayment of a loan --, discounted by the interest-rate d = (r - 1):
pv of constant-annuity = sum, from j=1 to j=i, of (p r^(-j)) = p (r^(-1) - r^(- (i + 1))) / (1 - r^(-1)) = p (1 - r^(-i)) / (r - 1) = p (1 - (1 + d)^(-i)) / d
limit, as j increases without bound, of the pv = p / d,
where the convergence radius of one on r translates to 0 < d. The discounting interest-rate d usually is taken as the
interest-rate upon a thirty-year US Treasury note.
Now, consider that the payments may be increasing by the compounded-rate (1 + b)^j -- for example, the property tax. Then, the geometric
series is:
pv of increasing-annuity = sum, from j=1 to j=i, of (p ((1 + b) / (1 + d))^j).
Hence, the effective interest-rate eff is
1 + eff = (1 + d) / (1 + b)
0 < eff = (1 + d) / (1 + b) - 1 = (d - b) / (1 + b)
limit, as j increases without bound, of the pv of the increasing-annuity = p (1 + b) / (d - b),
where the convergence criterion is that -1 < b < d. The b=0 takes us back to the constant-annuity. The stated eff
is to be substituted, into the formulae for the constant-annuity, in place of the d.
On the other hand, the payments may be decreasing by the reciprocal compounded-rate (1 + b)^(-j) -- for example, the depreciation of some asset,
like a building. Then, the geometric series is:
pv of decreasing-annuity = sum, from j=1 to j=i, of (p ((1 + b) (1 + d))^(-j)).
Hence, the effective interest-rate eff is
1 + eff = (1 + b) (1 + d)
0 < eff = b + d + b d
limit, as j increases without bound, of the pv of the decreasing-annuity = p / (b + d + b d).
The b=0 again takes us back to the constant-annuity.
We implicitly define the half-life hl by the first of the equations:
1 / 2 = (1 + b)^(- hl)
hl = ln(2) / ln(1 + b)
b = exp(ln(2) / hl) - 1.
For a property-tax upon a depreciating asset, we take the p as the current-tax. For the depreciation itself, we take the p as the principal P times the interest-rate b, namely p = P b.
The discounted; that is, present-value of these various annuities are
constant-annuity -- repayment of a loan or mortgage (of principal P) or the payments p of a loan, mortgage, or rent
pv of payments (constant-annuity) = (P b / (1 - (1 + b)^(- i))) ((1 - (1 + d)^(- i)) / d) --> P b / d
pv of constant-annuity = p (1 - (1 + d)^(- i)) / d --> p / d
increasing-annuity -- property tax upon land (take p as the current tax)
eff of incresing-annuity = (d - b) / (1 + b)
pv of increasing-annuity --> p (1 + b) / (d - b)
decreasing-annuity -- depreciation of an asset (take p = P b) or tax upon a depreciating asset (take p as the current tax)
b = exp(ln(2) / hl) - 1
eff of decreasing-annuity = b + d + b d
pv of decreasing-annuity --> p / (b + d + b d).
An Excel financial spread-sheet is available which discounts (capitalizes) the various expenses involved in a real-estate investment
and compares it to renting.
Enter the discount-rate into the cell D2. Ordinarily, the 30-year USA Treasury bond interest-rate is employed. Currently, this
interest-rate is between 4 and 4.5%. Historically, it has been between 6 and 7%. However, if you have significant investments in some
mutual funds, you may prefer to take its long-term yield, instead.
There are three examples provided, each consists of a column for the input-data (with a <<< flag), a column of intermediate results, and a
column for the output-data (with a >>> flag).
Enter a zero as the sum for any item which is not applicable. For easier readability, it is suggested that all money amount be scaled by the
same convenient multiple of a power of ten.
initial one-time:
closing costs <<<
down-payment <<<
increasing: property-tax
growth rate (%) [in the semi-closed interval [0, discount-rate).] <<<
current tax (yearly) [split the tax between here and on the depreciating asset.] <<<
discounted >>>
finite constant: mortgage
principal <<<
interest rate (%) [in the open interval (0, oo).] <<<
term (years) <<<
monthly payment
discounted >>>
infinite constant: utilities, insurance, maintenance, rent
yearly <<<
discounted >>>
decreasing: depreciation & property tax thereupon
half-life (years) [in the open interval (0, oo).] <<<
decay rate (%) <<<
current tax (yearly) [that portion upon the depreciating buildings. Enter a zero here and the whole tax in the "increasing" section; if the property tax does not take into account the depreciation of the buildings.] <<<
discounted >>>
asset [this would be the portion of the investment for the buildings.] <<<
discounted >>>
results
value of land
total capitalization >>>
net (less land) >>>
does it pay to refinance?
remaining balance: mortgage
interest rate (%) [in the open interval (0, oo).] <<<
term (years) <<<
monthly payment <<<
principal
discounted >>>
refinancing: mortgage
closing costs <<<
principal
interest rate (%) [in the open interval (0, oo).] <<<
term (years) <<<
tentative monthly payment
actual monthly payment <<<
amount financed
cash back >>>
discounted >>>
savings
A copy of this spread-sheet is available upon request mailto:finincial@rism.com Please ask for the financial spread-sheet.
The powers progressions often are called higher-degree arithmetic progressions. These progressions are of the form
a(i) = i^n
and the corresponding series is defined recursively as
A(i + 1) = A(i) + a(i + 1)
where, by the definition of series as the partial sums, A(0) = 0. For the term where both n and i are zero, we define zero to the zero power as one.
While for lower values of n, it is not too difficult to verify the closed forms of the series by Mathematical Induction, the derivation of the formulae requires the inversion of a matrix. It is obvious that the series is a polynomial of degree one higher, namely n + 1. The easiest algorithm is to fit a polynomial to the points [0, n + 1] and extrapolate.
While any interpolation formula may be employed and -- because of the uniqueness of the resulting polynomial -- the answer will be the same, since we have our own favorite interpolation formula, we will employ it. Then, in the notation for our formula for interpolation, we have
| n | g(u) | a(i) | z(ui) | A(i) |
|---|---|---|---|---|
| 0 | 2 u - 1 | 1, 1 | 1, 2 | i + 1 |
| 1 | u - 1 | 0, 1, 2 | 0, 1, 3 | i (i + 1) / 2 |
| 2 | 2 u - 3 | 0, 1, 4, 9 | 0, 1, 5, 14 | i (2 i^2 + 3 i + 1) / 6 |
| 3 | u - 2 | 0, 1, 8, 27, 81 | 0, 1, 10, 46, 146 | (i (i + 1) / 2)^2 |
| 4 | 2 u - 5 | 0, 1, 16, 81, 256, 375 | 0, 1, 17, 98, 354, 729 | i (6 i^4 + 15 i^3 + 10 i - 1) / 30 |
| 5 | u - 3 | 0, 1, 32, 243, 1024, 3275 | 0, 1, 33, 266, 1290, 4565 | i^2 (2 i^4 + 6 i^3 + 5 i^2 - 1) / 12 |
However, there is an easier way of obtaining these formulae.
Copyright © 1997, 8; 2003 R. I. 'Scibor-Marchocki last modified on Tuesday 14-th January 2003. Webmaster@rism.com.