我正在尝试在Python中实现差分方程。给定y 0 ，其中y 0 的形式为y n + 1 = a * y n + b是初始值并进行迭代-表示
y 1 = a * y 0 + b，
y 2 = a * y 1 + b，
...
一个示例问题（来自我的微积分类）将是： 假设您借了一笔60,000美元的贷款，并计划每月以1.2％的利率偿还700美元。五年后还剩下多少？将其设置为y n + 1 = 1.1 * y n -700，y 0 = 60,000
例如，您可以说，我理解Python中的递归
i = 0
while i < 20:
i = i+1
但是我不确定当下一次迭代需要上一次迭代的值时如何处理它。
答案 0 :(得分：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)
结果：$ 20444.98
或使用定义为sum of geometric progression的代数
Ln = S * r y -Pmt *（（1-r y ）/（1-r））= $ 20444.98
抱歉，MathJax不支持Stackoverflow。
答案 1 :(得分：1)
通常情况下，您可以通过使用计算中的最后一个值存储单个变量，然后将其作为初始值作为种子。 EG：
y = 60000
while True:
y = .1 * y - 700
当然，您必须确定何时停止以及如何处理这些值。您当然要打印它们：
y = 60000
while True:
y = .1 * y - 700
print(y)
但是您可能只想做100次左右，而不是永远做一次：
y = 60000
for i in range(12*5):
y = .1 * y - 700
print("%d: %f" % (i,y))
您可能还希望将它们存储起来以备后用，因此请将它们放在数组中：
y = 60000
results=[]
for i in range(12*5):
y = .1 * y - 700
results.append(y)
print(results)
答案 2 :(得分：0)
任务的常规循环如下所示：
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))
请注意，出于某些非特定于编程的原因，可能会讨论monthlyinterest：
a）它可能取决于何时计算利息，即每月的哪一天。可能在付款之前或之后。
b）对于银行来说，将年利率除以12更好。也可以将其计算为1.012的第12个根。
现在，上面是直接循环。它与递归无关。但可以将其转换为递归调用：
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)
关键是：
a）递归调用需要退出条件。我一开始就说对了。这里的退出条件是剩余的月份达到0时。
b）函数recursive()调用自身，希望它带有一个指向退出条件的参数。这里是month-1。
您这样称呼它：
print(recursive(60000, 1.2, 700, 60))
答案 3 :(得分：0)
按照您的要求使用递归：
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)
[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