Proof by Induction

kathrynag

1. The problem statement, all variables and given/known data

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

2. Relevant equations



3. 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

mutton

Use this:

kathrynag

I don't really understand how I use that...

mutton

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.

kathrynag

Ok, so:
(1+x)^k(1+x)>(1+kx)(1+x)
>(1+2kx+x)

mutton

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?

kathrynag

No restrictions.
1+kx+x+kx^2

mutton

When k = 3 and x = -4,

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

which makes the statement false.

kathrynag

So, we have to assume 1+x>0

kathrynag

so, x>-1

Mark44

I believe x has to be strictly within 1 unit of 1; i.e., |1 + x| < 1, which means that 0 < x < 2.

kathrynag

Once I get here I'm unsure where to go

mutton

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