Prove P(n): "Proof by Induction for (1+h)n\geq1+nh+\frac{n(n+1)}{2}h2

  • Thread starter Thread starter major_maths
  • Start date Start date
  • Tags Tags
    Induction Proof
Click For Summary

Homework Help Overview

The problem involves proving the inequality (1+h)^n ≥ 1 + nh + (n(n+1)/2)h^2 for any positive h and integer n ≥ 0, using mathematical induction.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the base case and the inductive step, with one participant attempting to relate the expression for P(K+1) to P(K). Others are questioning the terms used in their inductive hypothesis, particularly the difference between k(k-1) and k(k+1).

Discussion Status

There is ongoing exploration of the inductive hypothesis and its implications. Some participants are providing guidance on how to manipulate the expressions, while others are clarifying their understanding of the terms involved in the proof.

Contextual Notes

Participants note discrepancies in their notes regarding the terms used in the inductive hypothesis, leading to confusion about the correct formulation. There is also a reference to specific values (e.g., n=2) to illustrate the point being discussed.

major_maths
Messages
30
Reaction score
0

Homework Statement


Prove that for any positive h and any integer n\geq0, (1+h)n\geq1+nh+\frac{n(n+1)}{2}h2.


Homework Equations


None.


The Attempt at a Solution


I proved that P(0) is true (1\geq1). The rest of the proof goes as follows:

Assume K\inZ (the set of integers) and P(K) is true.
Then (1+h)K\geq1+Kh+\frac{K(K-1)}{2}h2.
Then (1+h)(K+1) = (1+h)K+(1+h)1...

I can't figure out how to relate that part to the final part of P(K+1), which is 1+(K+1)h+\frac{(K+1)(K+1-1)}{2}h2.
 
Physics news on Phys.org
You have that (1+h)K+1=(1+h)K(1+h) = P(K)(1+h), so see where that goes.
 
I'm working on the same problem, to show P(k+1) I set it up the same way

but then we can use our inductive hypothesis

(1+x)^(k+1) >= (1+kx+(1/2)*k(k-1)*x^2)(1+x)

My question is, I've wrestled with the algebra for a little while now and for some reason in my notes i had P(K) set to:

1+kx+(1/2)*k(k+1)*x^2

(where the 1 in k(k+1) is positive instead of negative) I think my professor did the problem with k+1 instead of k-1. But i thought from the inductive hypothesis the term is k(k-1) NOT k(k+1) because k+1 is what we get from P(k+1) that is what we get from the substitution?

I know there will be left over terms but i keep getting k(k-1)/2 instead of k(k+1)/2.
 
stihl29 said:
I'm working on the same problem, to show P(k+1) I set it up the same way

but then we can use our inductive hypothesis

(1+x)^(k+1) >= (1+kx+(1/2)*k(k-1)*x^2)(1+x)

My question is, I've wrestled with the algebra for a little while now and for some reason in my notes i had P(K) set to:

1+kx+(1/2)*k(k+1)*x^2

(where the 1 in k(k+1) is positive instead of negative) I think my professor did the problem with k+1 instead of k-1. But i thought from the inductive hypothesis the term is k(k-1) NOT k(k+1) because k+1 is what we get from P(k+1) that is what we get from the substitution?

I know there will be left over terms but i keep getting k(k-1)/2 instead of k(k+1)/2.

I believe you are correct. It has to be n(n-1) instead of n(n+1).

This is seen by picking n=2, then

(1+h)^2=1+2h+h^2

and not

(1+h)^2=1+2h+3h^2
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
9
Views
2K
  • · Replies 23 ·
Replies
23
Views
2K
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K