在 Python 中实现差分（非差分）方程

Question

I'm trying to implement a difference equation in Python. These take the form y _n+1 = a * y _n + b, given y ₀ where y ₀ is initial value and it iterates -- meaning

y ₁ = a * y ₀ + b,
y ₂ = a * y ₁ + b,
...

An example problem (from my calculus class) would be thus: Say you take out a loan of $60,000 and plan on paying $700 a month back at an interest rate of 1.2%. How much would be left after 5 years? This would be set up as y _n+1 = 1.1 * y _n - 700, y ₀ = 60,000

I understand recursion in Python in the sense that you can say, for example,

i = 0
while i < 20:
    i = i+1

but I'm unsure how to approach it when the next iteration requires a value from the previous.

Answer 1

def calc1():
    S=60000
    x=0.012
    y=5*12
    Pmt=700
    r=1+(x/12)
    Ln = S
    for k in range(1, y+1):
         Ln=(Ln*r)-Pmt
    print (Ln)

Result: $20444.98

Or using algebra with definition of sum of geometric progression

Ln= S * r ^y - Pmt * ((1-r ^y ) / (1-r)) = $20444.98

Sorry Stackoverflow is not supported MathJax.

Answer 2

Typically you'd do this by storing a single variable with the last value from the calculation, and seeding it with your starting value. EG:

y = 60000
while True:
    y = .1 * y - 700

Of course, you have to determine when to stop and what to do with the values. You certainly want to print them:

y = 60000
while True:
    y = .1 * y - 700
    print(y)

But you probably only want to do it for 100 times or something instead of forever:

y = 60000
for i in range(12*5):
    y = .1 * y - 700
    print("%d: %f" % (i,y))

And you may want to store them for use later as well, so put those in an array:

y = 60000
results=[]
for i in range(12*5):
    y = .1 * y - 700
    results.append(y)

print(results)

Answer 3

A regular loop for your task would look like this:

loan = 60000
annualinterest = 1.2/100
monthlyinterest = annualinterest/12
monthlyfactor = 1+monthlyinterest
monthlypackback = 700

for month in range(5*12):
    loan = loan * monthlyfactor - monthlypackback
    print(month+1, "{0:.2f}".format(loan))

Note that the monthlyinterest might be discussed for several reasons which are not programming specific:

a) it may depend when the interest is calculated, ie on which day of the month. That may be before or after the payment.

b) dividing the annual rate by 12 is better for the banks. It could also be calculated as the 12th root of 1.012.

Now, above is a straight forward loop. It has nothing to do with recursion. But it can be converted into a recursive call:

def recursive(loan, annualinterestPercentage, monthlyPayback, months):
    if months == 0:
        return loan
    annualinterest = annualinterestPercentage / 100
    monthlyinterest = annualinterest / 12
    monthlyfactor = 1 + monthlyinterest
    return recursive(loan * monthlyfactor - monthlyPayback, annualinterestPercentage, monthlyPayback, months - 1)

The key things are:

a) a recursive call needs an exit condition. I put it right at the beginning. The exit condition here is when the remaining months reach 0.

b) the function recursive() calls itself, hopefully with a parameter that goes towards the exit condition. Here it's month-1 .

You call it like this:

print(recursive(60000, 1.2, 700, 60))

Answer 4

Code

Using recursion as you asked:

def loan_calculator(y, payments, interest, monthly, period, counter=0, amount_left=[]):
  # y -> starting loan (60,000) 
  # payments -> paying $700/month
  # interest -> Interest rate on loan (yearly)
  # monthly -> boolean, if monthly or yearly
  # period -> how many years
  # counter -> number of payments made
  # amount_left -> after each payment how much is left to pay

  if monthly:
    period *= 12
    interest /= (12*100)

  # stopping condition:
  if counter == period:
    return y, amount_left
  else:
    y = (1+interest)*y - payments
    amount_left.append(y)
    counter += 1
    return loan_calculator(y, 700, 1.2, True, 5, counter)

final_payment, all_payments = loan_calculator(6e4, 700, 1.2, True, 5)
print(all_payments)
print(final_payment)

Output

[59359.99999999999, 58719.359999999986, 58078.07935999998, 57436.15743935997, 56793.59359679933, 5615
0.38719039612, 55506.53757758651, 54862.04411516409, 54216.90615927925, 53571.12306543852, 52924.6941
88503956, 52277.618882692455, 51629.896501575146, 50981.52639807671, 50332.50792447478, 49682.8404323
9925, 49032.52327283165, 48381.555796104476, 47729.937351900575, 47077.66728925247, 46424.74495654172
, 45771.16970149826, 45116.94087119975, 44462.05781207095, 43806.51986988301, 43150.32638975289, 4249
3.476716142635, 41835.970192858775, 41177.80616305163, 40518.98396921468, 39859.50295318389, 39199.36
245613707, 38538.561818593196, 37877.100380411786, 37214.97748079219, 36552.192458272984, 35888.74465
0731256, 35224.633395381985, 34559.858028777366, 33894.41788680614, 33228.31230469295, 32561.54061699
7634, 31894.102157614627, 31225.996259772237, 30557.222256032004, 29887.779478288034, 29217.667257766
32, 28546.884925024082, 27875.431809949103, 27203.30724175905, 26530.510549000803, 25857.0410595498, 
25182.898100609345, 24508.080998709953, 23832.58907970866, 23156.421668788367, 22479.578090457155, 21
802.057668547608, 21123.859726216153, 20444.983585942366] # This is a list of payments you owe each month                                           
20444.983585942366 # This is your final payment