Prove x^n > x for all x > 1.

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November 20th 2009, 05:54 PM
450081592

Prove x^n > x for all x > 1.

Suppose n > 1 is a positive integer. Prove $x^n > x$ for all x > 1.
(hint: suppose we know f and g are differentiable on the interval (-c,c) and f(0) = g(0), if f'(x) > g'(x) for all $x \in(0,c)$ , then f(x) > g(x) for all $x \in (0,c)$ .
November 20th 2009, 06:38 PM
lvleph

Do you know how to prove something by induction? If not I am unsure how to do this for general n.
November 20th 2009, 06:46 PM
450081592

Quote:

Originally Posted by lvleph

Do you know how to prove something by induction? If not I am unsure how to do this for general n.

we have done a couple of induction prove, not so familiar with that, can we apply the thereom in the hints that is given?
November 20th 2009, 06:49 PM
lvleph

What you want to do is use the hint on a base case (n=2). Then you will make the assumption that the problem hypothesis holds for n=k. Use this assumption to prove that it is true for n=k+1.
November 20th 2009, 06:52 PM
450081592

Quote:

Originally Posted by lvleph

What you want to do is use the hint on a base case (n=2). Then you will make the assumption that the problem hypothesis holds for n=k. Use this assumption to prove that it is true for n=k+1.

why do we set n = 2?
November 20th 2009, 06:54 PM
lvleph

Because it is easy and the problem statement says n > 1.
November 20th 2009, 06:59 PM
450081592

Quote:

Originally Posted by lvleph

Because it is easy and the problem statement says n > 1.

then how are we gonna apply the hint to the proof, I can't see the relation between them
November 20th 2009, 07:03 PM
Drexel28

Quote:

Originally Posted by 450081592

Suppose n > 1 is a positive integer. Prove $x^n > x$ for all x > 1.
(hint: suppose we know f and g are differentiable on the interval (-c,c) and f(0) = g(0), if f'(x) > g'(x) for all $x \in(0,c)$ , then f(x) > g(x) for all $x \in (0,c)$ .

Note that if $f(x)=x^n\quad n>1$ and $g(x)=x$ . That $f(1)=1=g(1)$ and $f'(x)=nx^{n-1}\ge n>1=g'(1)$ and the concluso follows.
November 20th 2009, 07:05 PM
lvleph

Okay, I will show you the base case.

Assume n=2.
$f(x) = x^2, \; g(x) = x$
Now let's show that these are equal at $x=0$
$f(1)=1^2 = 1 = g(1)$
Now we differentiate the two
$f'(x) = 2x$
$<br /> g'(x) = 1$
For $x>1$
$<br /> 2x > 2$
Therefore we see that $f'(x)=2x>1 =g'(x)$ for $x>1$ .
November 20th 2009, 07:07 PM
lvleph

Quote:

Originally Posted by Drexel28

Note that if $f(x)=x^n\quad n>1$ and $g(x)=x$ . That $f(1)=1=g(1)$ and $f'(x)=nx^{n-1}>n>1=g'(1)$ and the concluso follows.

In fact, I guess this is a much quicker and simpler proof.
November 20th 2009, 08:30 PM
450081592

Quote:

Originally Posted by lvleph

In fact, I guess this is a much quicker and simpler proof.

so with this method, we do not need to prove by induction, right?
November 20th 2009, 09:26 PM
450081592

Quote:

Originally Posted by Drexel28

Note that if $f(x)=x^n\quad n>1$ and $g(x)=x$ . That $f(1)=1=g(1)$ and $f'(x)=nx^{n-1}\ge n>1=g'(1)$ and the concluso follows.

how do we know g(x) is x?

That $f(1)=1=g(1)$ and $f'(x)=nx^{n-1}\ge n>1=g'(1)$

and I dont get this line
November 20th 2009, 09:33 PM
lvleph

We know $g(x)=x$ because you are trying to prove $x^n > x$ and the hint was suggesting you $f(x) > g(x)$
Wouldn't you agree $nx^{n-1} \ge n$ for $x>1$ and $n>1$ ? But we know that $f'(x) = nx^{n-1}$ and $g'(x) = 1$ .
November 20th 2009, 09:35 PM
450081592

Quote:

Originally Posted by lvleph

We know $g(x)=x$ because you are trying to prove $x^n > x$ and the hint was suggesting you $f(x) > g(x)$
Wouldn't you agree $nx^{n-1} \ge n$ for $x>1$ and $n>1$ ? But we know that $f'(x) = nx^{n-1}$ and $g'(x) = 1$ .

so we are just assuming g(x) = x?
November 20th 2009, 09:37 PM
lvleph

Yes, because the hint was to prove $f(x) > g(x)$ and we wanted to prove $x^n > x$ so it makes sense to assume $g(x)=x$ .

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