Mathematical Induction problem. (1 Viewer)

[ascentyx]

Prove 2^n > n^3 for n>9.

I can do this geometrically but i want to know how to do it the 'proper' way. Yes i have finished my hsc, but i haven't done this for ages so i have forget everything haha.

 

[Timothy.Siu]

Prove 2^n > n^3 for n>9.

I can do this geometrically but i want to know how to do it the 'proper' way. Yes i have finished my hsc, but i haven't done this for ages so i have forget everything haha.

Prove 2^n > n^3 for n>9.
i.e. Prove 2^n-n^3>0
test for n=10,

LHS= 1024-1000=24 >0, therefore n=10 is true

asssume true for some particular n=k,
i.e, 2^k-k^3>0

test for n=k+1
i.e. prove 2^(k+1)-(k+1)^3 >0
LHS=2.2^k-(k+1)^3
=2(2^k-k^3)+2k^3-(k+1)^3 (i just added -2k^3+2k^3 = 0)
>2k^3-(k^3+3k^2+3k+1) (subbed in the n=k)
=k^3-3k^2-3k-1=(k-1)^3+6k which is clearly >0 for k>9

 

[ascentyx]

Prove 2^n > n^3 for n>9.
i.e. Prove 2^n-n^3>0
test for n=10,

LHS= 1024-1000=24 >0, therefore n=10 is true

asssume true for some particular n=k,
i.e, 2^k-k^3>0

test for n=k+1
i.e. prove 2^(k+1)-(k+1)^3 >0
LHS=2.2^k-(k+1)^3
=2(2^k-k^3)+2k^3-(k+1)^3 (i just added -2k^3+2k^3 = 0)
>2k^3-(k^3+3k^2+3k+1) (subbed in the n=k)
=k^3-3k^2-3k-1=(k-1)^3+6k which is clearly >0 for k>9

I tried to do exactly what you did but failed. Correct me if i'm wrong but with the last line, doesn't (k-1)^3 = k^3 - 3k^2 + 3k -1 so then wouldn't

k^3-3k^2-3k-1=(k-1)^3-6k and then the problem arises again

 

[tommykins]

(k-1)^3 > 0 as k>9 so the smallest value is 8^3 for that expression.

 

[ascentyx]

(k-1)^3 > 0 as k>9 so the smallest value is 8^3 for that expression.

I know that but how can i prove it's greater than 6k.

 

[Timothy.Siu]

I know that but how can i prove it's greater than 6k.

yeah my bad, careless, but i'm pretty sure u can just say

(k-1)^3-6k > 0 for k>9 (obvious)

or u can just draw it

 

[tommykins]

oh i thought it was a +6k.

and no, if its -6k you have to prove (k-1)^3 > 6k

 

[Timothy.Siu]

oh i thought it was a +6k.

and no, if its -6k you have to prove (k-1)^3 > 6k

yeah...i guess so, so either u can draw it, or say, (k-1)^3 increases faster than 6k at starts bigger(k=10) , which cud be done by finding the derivate

 

[ascentyx]

yeah...i guess so, so either u can draw it, or say, (k-1)^3 increases faster than 6k at starts bigger(k=10) , which cud be done by finding the derivate

Yeah i did it by drawing it. I was wondering if there was a better way.

 

[tommykins]

you can try proving (k-1)^3 - 6k > 0 for k > 9 and also prove that the differential is > 0 for all k>9

 

[Iruka]

It's an inequality so you can do a bit of sloppy maths:

We want to show that if 2^k>k^3, then 2^(k+1)> (k+1)^3

So suppose that 2^k > k^3, and that k>9.

Then 2^(k+1) = 2.2^k > 2k^3 = k^3+k^3
If we can show that k^3> 3k^2+3k+1, then we can say that

2k^3 > k^3+3k^2+3k+1 = (k+1)^3, and the required inequality holds.

Since k>9, k^3>9k^2

And since k>1, 9k^2= 3k^2+3k^2+3k^2>3k^2+3k+1

Thus k^3> 3k^2+3k+1, as required.

 

[Pwnage101]

yeh this Q from MATH132 gave me a bit of trouble as well

 

[jet]

You can either, a) plug the value in and establish that a cubic will be larger if it starts off larger [larger than a linear]
or b) differentiate f(k) = (k - 1)^3 - 6k, i.e f'(k) = 3(k - 1)^2 > 0 for all k, therefore, (k - 1)^3 - 6k increases for all k. Therefore, since k>9, f(9) = 16 (i hope, im doing this in my head), then f(x)>0 for all k.