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

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Nov 20th 2009, 06:54 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 $x \in (0,c)$.
• Nov 20th 2009, 07:38 PM
lvleph
Do you know how to prove something by induction? If not I am unsure how to do this for general n.
• Nov 20th 2009, 07:46 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?
• Nov 20th 2009, 07:49 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.
• Nov 20th 2009, 07:52 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?
• Nov 20th 2009, 07:54 PM
lvleph
Because it is easy and the problem statement says n > 1.
• Nov 20th 2009, 07:59 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
• Nov 20th 2009, 08:03 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.
• Nov 20th 2009, 08:05 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
$f'(x) = 2x$
$
g'(x) = 1$

For $x>1$
$
2x > 2$

Therefore we see that $f'(x)=2x>1 =g'(x)$ for $x>1$.
• Nov 20th 2009, 08:07 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.
• Nov 20th 2009, 09:30 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?
• Nov 20th 2009, 10:26 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
• Nov 20th 2009, 10:33 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 $g'(x) = 1$.
• Nov 20th 2009, 10:35 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?
• Nov 20th 2009, 10:37 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 $g(x)=x$.
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last