Proof by Induction

  • Thread starter kathrynag
  • Start date
  • #1
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Homework Statement



[tex](1+x)^{k}[/tex][tex]\geq[/tex]1+kx

Homework Equations





The Attempt at a Solution


I want to show for P(k+1)
(1+x)^(k+1)[tex]\geq[/tex]1+kx+x
(1+x)^k*(1+x)[tex]\geq[/tex]1+kx+x
 

Answers and Replies

  • #2
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(1+x)^k*(1+x)[tex]\geq[/tex]1+kx+x

Use this:

[tex](1+x)^{k}[/tex][tex]\geq[/tex]1+kx
 
  • #3
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I don't really understand how I use that...
 
  • #4
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You want to use an inequality involving [tex](1 + x)^k[/tex] to derive an inequality involving [tex](1+x)^k (1+x)[/tex], which is a multiple of it.

An inequality still holds if you multiply both sides by a positive number. If you multiply both sides of an inequality by a negative number, then you have to flip the sign from [tex]\le[/tex] to [tex]\ge[/tex] or vice versa.
 
  • #5
598
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Ok, so:
(1+x)^k(1+x)>(1+kx)(1+x)
>(1+2kx+x)
 
  • #6
179
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Right idea, but there is an error in your expansion.

Does the question give any restrictions on x? Is what you wrote still true of 1 + x < 0?
 
  • #7
598
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No restrictions.
1+kx+x+kx^2
 
  • #8
179
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When k = 3 and x = -4,

[tex](1 + x)^k = (-3)^3 = -27 < -11 = 1 - 12 = 1 + kx[/tex]

which makes the statement false.
 
  • #9
598
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So, we have to assume 1+x>0
 
  • #10
598
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so, x>-1
 
  • #11
35,237
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So, we have to assume 1+x>0
I believe x has to be strictly within 1 unit of 1; i.e., |1 + x| < 1, which means that 0 < x < 2.
 
  • #12
598
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No restrictions.
1+kx+x+kx^2

Once I get here I'm unsure where to go
 
  • #13
179
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Once I get here I'm unsure where to go

Look at your first post; you need to show that that is [tex]\ge 1 + kx + x[/tex].
 

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