Proof by Induction: P(k) to P(k+1)

[kathrynag]

Homework Statement

$$(1+x)^{k}$$$$\geq$$1+kx

Homework Equations

The Attempt at a Solution

I want to show for P(k+1)
(1+x)^(k+1)$$\geq$$1+kx+x
(1+x)^k*(1+x)$$\geq$$1+kx+x

 

[mutton]

(1+x)^k*(1+x)$$\geq$$1+kx+x

Use this:

$$(1+x)^{k}$$$$\geq$$1+kx

 

[kathrynag]

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

 

[mutton]

You want to use an inequality involving $$(1 + x)^k$$ to derive an inequality involving $$(1+x)^k (1+x)$$, 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 $$\le$$ to $$\ge$$ 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,

$$(1 + x)^k = (-3)^3 = -27 < -11 = 1 - 12 = 1 + kx$$

which makes the statement false.

 

[kathrynag]

So, we have to assume 1+x>0

 

[kathrynag]

so, x>-1

 

[Mark44]

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.

 

[kathrynag]

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

Once I get here I'm unsure where to go

 

[mutton]

Once I get here I'm unsure where to go

Look at your first post; you need to show that that is $$\ge 1 + kx + x$$.