Finding N: Logarithmic Approach Needed

Click For Summary

Homework Help Overview

The problem involves finding a value of N in the context of an inequality involving factorials and powers of pi and 2. The expression is given as \(\frac{\pi^{(N+1)}}{(N+1)!} \cdot \frac{1}{2^N} \leq 10^{-5}\).

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the potential use of logarithms to simplify the problem and evaluate the expression for various values of N. There is uncertainty about whether to find any N or the smallest such N, and some participants suggest trying integer values for N.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and suggesting various approaches to evaluate the expression. Some guidance has been offered regarding estimating values and evaluating the expression for different N.

Contextual Notes

There is a mention of the expression decreasing with increasing N, and the challenge of estimating log(n!) without leading to a clear solution. Participants are also considering the implications of the inequality in terms of finding a suitable N.

Fairy111
Messages
72
Reaction score
0

Homework Statement



I need to find N:

{[pi^(N+1)] / (N+1)!} . {1/2^N} <= 10^-5

Homework Equations





The Attempt at a Solution



I'm really struggling with this, i think taking logs is involved, but i can't seem to find a value of N.

Any help?
 
Physics news on Phys.org
The expression decreases with increasing N. You basically want to evaluate it for various values of N until you figure out where it starts going below 10^(-5). You can only estimate log(n!) and the estimate doesn't lead to anything you can really solve.
 
Do you need to find any N, or the smallest such N?
 
I need to find what N needs to be bigger or equal to.
 
Then just try some integer values for N. N doesn't have to be very big.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K