precal question

Question

Find and prove an inequality relating 100n and n³.

Since 100n

 and n³

 for n = 1, 2, 3, . . . 9, 10, 11

 are 110, 200, 300, . . . 900, 1000, 1100

 and 1, 8, 37, . . . 729, 1000, 1331

 respectively, our conjecture is that 100n ≤ n³,

 for all n ≥  .

 Let P(n)

 denote the statement that 100n ≤ n³.

P(10)

is the statement that 100		≤		3

,

 which is true.

Assume that P(k)

 is true. Thus, our induction hypothesis is 100k  ? 

 for some k ≥  .

We want to use this to show that P(k + 1)

 is true. Now,

100(k + 1) = 100k +  ≤ k + k² ≤ k³ + 3k² + 3k + 1 =

 

 

	3

.

Thus, P(k + 1)

 follows from P(k),

 and this completes the induction step. Having proven the above steps, we conclude by the Principle of Mathematical Induction that P(n)

 is true.

Answer 1

The conjecture is that 100n <= n^3 for n >= 11

For n = 11, 100n = 1000 and 11^3 = 1331

1000 <=1331 is True

Assume for n =k

For n=k+1

100(k+1) - 100k =100

k+1^3 -k^3 = k^3 +3k^2 + 3k +1 - k^3 =3k^2 +3k + 1

3k^2 +3k + 1 - (3(k-1)^2 +3(k-1))+1) =

3k^2 +3k + 1 - (3k^2 -6k + 3 +3k - 3 + 1) =

6k >= 0 for k>= 1, so the difference is nondecreasing for k >= 11

For k = 11, 3k^2 +3k + 1 =3*11^2 +3*11 + 1 = 397>=100

Thus, as

100k <= k^3 and

100 <= 3k^2 +3k +1 for k >= 11

100k+100 <= k^3 + 3k^2 +3k +1 or

100(k+1) <= (k+1)^3 for k >= 11

We have proven for k=11 and proven that, if true for k>=11, it is true for k+1, so

100n <= n^3 for n >= 11

Answer 2

Hi Viviana C.,

Most of this question is a proof completion, which is already supplied to you?

So what is the problem with noting that 100n=n^3 for n=10, and showing the inequality for n>10? (Note that that's not limited to the integers, even). You wouldn't even need induction for that?

I guess you would have to also include the "opposite" inequality for n<-10, unless your domain was explicity only the positive integers.

-- Cheers, --Mr. d.