<no subject>

[HGL]

Hello Everyone-- I need a little help proving 2^n > n^2 for all n > 4
Could someone give me some hints to continue after:
2^(k+1) > (k+1)^2
2*2^k > k^2 + 2k + 1

Thanks

[William Elliot]

Post by HGL
Hello Everyone-- I need a little help proving 2^n > n^2 for all n > 4
2^(k+1) > (k+1)^2
2*2^k > k^2 + 2k + 1

How fortuitous. I'm a mind reader. You are using induction.
Why are you using induction? Should you be using deduction?

Ok, just to test your psychic skills
2^k + 2^k > k^2 + 2k + 1

[Nicholas Sherlock]

Post by HGL
Hello Everyone-- I need a little help proving 2^n > n^2 for all n > 4
2^(k+1) > (k+1)^2

Assuming that you're doing an induction here...

Okay,
(k+1)^2 <= k^2 + 2k + 1
<= 2^k + 2k + 1 (by I.H.)
<= 2^k + 2k + k (since k>=1)
<= 2^k + 3k
<= 2^k + k^2 (since k>=3)
<= 2^k + 2^k (by I.H.)
<= 2^(k+1)

As req'd.

Cheers,
Nicholas Sherlock

[Noehl]

Post by HGL
Hello Everyone-- I need a little help proving 2^n > n^2 for all n > 4
2^(k+1) > (k+1)^2
2*2^k > k^2 + 2k + 1
Thanks

try applying logarithms to both sides of the inequality
--
Noehl u144241

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[William Elliot]

Post by Noehl

Post by HGL
Hello Everyone-- I need a little help proving 2^n > n^2 for all n > 4
2^(k+1) > (k+1)^2
2*2^k > k^2 + 2k + 1

try applying logarithms to both sides of the inequality

n.log 2 > 2.log n

What now?

[Noehl]

Post by William Elliot

Post by Noehl

Post by HGL
Hello Everyone-- I need a little help proving 2^n > n^2 for all n > 4
2^(k+1) > (k+1)^2
2*2^k > k^2 + 2k + 1

try applying logarithms to both sides of the inequality

n.log 2 > 2.log n
What now?

prove that

n > 2 (log n)/(log 2)

which will always be true for all n > 4
--
Vempire u144241

Games that I like to play

Multiplayer Online Games <http://www.gamestotal.com/>
Strategy Games <http://www.gamestotal.com/>
Unification Wars <http://uc.gamestotal.com/>
Massive Multiplayer Online Games <http://uc.gamestotal.com/>
Galactic Conquest <http://gc.gamestotal.com/>
Strategy Games <http://gc.gamestotal.com/>
Runescape <http://www.stephenyong.com/runescape.htm>
Kings of chaos <http://www.stephenyong.com/kingsofchaos.htm>