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How can I compute the sum of P (n) with a recursive function for any depth n.

QUESTION

lumi
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animoxl
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    Recursion has 2 parts: the recursive condition (or recursive call) and the base (or stopping) condition. What have you tried so far? Have you analyzed/identified those 2 conditions for your example? – Ralf Mar 25 '21 at 20:02
  • The basic recursion pattern is: create function signature, test stopping condition and return, if exit condition not met perform intermediate step and pass value to recursive function. Take a look at recursive Fibonacci as an example of a recursive Python method that computes a recurrence relationship: https://stackoverflow.com/questions/494594/how-to-write-the-fibonacci-sequence – Michael Ruth Mar 25 '21 at 20:02
  • Start off with your exit condition as `n == 1`, and `return n**-2 + n**-4 + n**-6`. Then figure out how to perform the summation for n > 1. – Michael Ruth Mar 25 '21 at 20:08
  • @MichaelRuth Yeah. I already saw that fibonacci but this one is kinda confusing to me. It might because of the way it looks. Thanks a lot tho i will check it. Pretty sure it will help me! – animoxl Mar 25 '21 at 20:11
  • @animoxl, SO expects you to at least try something and post your code. This is why I throw out a bread crumb and a link rather than posting a solution. Give it a shot, post your output vs your expected output, and the SO community will help you out. – Michael Ruth Mar 25 '21 at 20:30

3 Answers3

0

Here's how you could do it:

def P(n, k = 1, currentSum = 0):
    if k > n:
        return currentSum
    currentSum += 1/k**2 + 1/k**4 + 1/k**6
    return P(n, k+1, currentSum)
    
print(P(3)) # 3.4529535322359397
lnogueir
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0

Is something as simple as:

def P(n):
    return n if n < 1 else P(n - 1) + 1/n**2 + 1/n**4 + 1/n**6

satisfactory to your needs?

cdlane
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0

A proposal for a longer but simpler(I think) solution(with an explanation):
Because of the commutative property(a+b = b+a), you can do things backwards. This means that you can solve this equation backwards and instead of incrementing n by 1, you can start with n and decrement by 1, adding as you go. This turns out to be perfect for recursion, as all you need to do is add the terms(1/n**2 and such) with the function for a slightly smaller number, with a corner case:

def P(n):
  if n == 0:
    return 0
  t1 = n**-2#** is the python syntax for raising powers
  t2 = n**-4
  t3 = n**-6
  sum_of_terms = t1+t2+t3
  return sum_of_terms + P(n-1)
print(P(100))
#3.7346498674943076

How it works:(calculating P(3))

 1. Start with 1/9 + 1/81 + 1/729 
 2. Add that to P(2): 
    a. Start with 1/4 + 1/16 + 1/64
    b. Add that to P(1)
        i. Start with 1/1 + 1/1 + 1/1
        ii. Add that to P(0)
        iii. To prevent division by zero, return 0
    c. So the sum of that is 3
 3. So the sum  of that is 3.328125
So the total is 3.4529535322359397

Note: Python's floats only hold so much precision.

NRO
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