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Inequality problem

  • Thread starter loli12
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loli12

Hi, I was asked to determine which values of n, 3^n > n^3.

I think this inequality holds for all intergers. But I have trouble proving it..

I checked n=1 which the inequality holds, then assume true for n.
for n+1 case:
3^(n+1) > (n+1)^3
3*3^n > (n+1)^3

i know that 3*3^n > 3n^3. But is there a way to show that 3n^3 > (n+1)^3?
and also, even i proved that to be right, it's only valid for n>= 4, how should i show that for n< 4 ?

Please give me some idea for that! Thanks
 
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loli12 said:
I think this inequality holds for all intergers. But I have trouble proving it..
I checked n=1 which the inequality holds, then assume true for n.
for n+1 case:
3^(n+1) > (n+1)^3
3*3^n > (n+1)^3
...But is there a way to show that 3n^3 > (n+1)^3?
hmmm...O(n) problem...i'm kinda rusty on this, but here's what i remember:

in this case, on the RHS you have n^3 term (if you expand the cube of sum), you can "ignore" the rest of the expression since the order of the polynomial is 3 here and everything else is going to be either on order of n^3 or less than that. So, what you'll end up doing is comparing 3^n and n^3 again (ommiting the constants) and the way induction works, you have reduced the inductive case (n+1) to the base case (which i assume you would have proved with n >=4), so ... end of proof.
For the latter (n >= 4) i will have to dig out my notes and get back to you, i remember doing it but don't remember how to do it formally (perhaps ln or smth like that). It's basically finding a point of intersection of these two graphs, since the obvious case is when n = 3, 3^n = n^3, i.e. have a common point.
 
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Usually these types of problems are difficult to solve because the variable is in both the base and the exponent. I don't know of any other way besides numerically.
 
Assume [itex]3^p > p^3[/itex] for p >= 4

Then [tex]3^{p+1} - (p+1)^3 = 3*3^p - (p+1)^3 > 3p^3 - (p+1)^3 = 3p^3 - p^3 - 3p^2 - 3p - 1 = 2p^3 - 3p^2 -3 p - 1[/tex].
Now just show that [itex]2p^3 - 3p^2 - 3p - 1 > 0[/itex] for all p >= 4.
 
Last edited:

Hurkyl

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how should i show that for n< 4 ?
By plugging in each leftover value in turn and checking the inequality. :tongue2:
 
L

loli12

Thanks all for replying!
i was able to show this to be true for all R <= 0 and all R >= 4. Next, i just plug in the value for n = 1, 2 which makes the inequality true. (But is it enough to assume that the Real numbers n, 0<= n < 3 satisifies the inequality?)
Besides for n=3, both sides are equal which makes the inequality false. then how should i show the Real numbers between 3 and 4 to be true without going into Caluculus and just by just using the induction method?


edit: oh, if i checked n=1 to be correct, for induction method, can i say that i assume all n to be true except case n=3. and when doing the inequality, apply log on both sides?
 
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