Mathematical Induction: How did that happen (underlined) (1 Viewer)

[Tryingtodowell]

 

[Tryingtodowell]

How did they go from = k^3 - (3k^2 +3k+1) to >k^3 - (3k^2 +3k^2+3k^2) ??????

 

[liamkk112]

lets quickly see that k^3 - (3k^2 + 3k + 1) > k^3 - (3k^2 + 3k^2 + 3k^2) since this is what they used
notice that we can cross out the k^3 on both sides and move the negative terms to the other side giving:
9k^2 > 3k^2 + 3k + 1
or that 6k^2 - 3k - 1 > 0
using the quadratic formula we can factorise:
(k - ((9 + sqrt(33))/12)(k - ((9 - sqrt(33))/12 > 0

and we can see that this inequality holds true, if k > 1

so we now can see that k^3 - (3k^2 + 3k + 1) > k^3 - (3k^2 + 3k^2 + 3k^2)

we can also think of it this way:
the LHS > RHS if the term in the brackets on the RHS is bigger than the term in the brackets on the LHS, as we should be subtracting more on the RHS to make it lesser than the LHS
we can clearly see that 3k^2 > 3k, 3k^2 > 1 as k > 1
so we know that what is in the brackets on the RHS is bigger than what is in the brackets on the LHS

hence the result on the RHS must be smaller than the result on the LHS, as we are subtracting more from the k^3 term on the RHS than the LHS

 

[Tryingtodowell]

lets quickly see that k^3 - (3k^2 + 3k + 1) > k^3 - (3k^2 + 3k^2 + 3k^2) since this is what they used
notice that we can cross out the k^3 on both sides and move the negative terms to the other side giving:
9k^2 > 3k^2 + 3k + 1
or that 6k^2 - 3k - 1 > 0
using the quadratic formula we can factorise:
(k - ((9 + sqrt(33))/12)(k - ((9 - sqrt(33))/12 > 0

and we can see that this inequality holds true, if k > 1

so we now can see that k^3 - (3k^2 + 3k + 1) > k^3 - (3k^2 + 3k^2 + 3k^2)

we can also think of it this way:
the LHS > RHS if the term in the brackets on the RHS is bigger than the term in the brackets on the LHS, as we should be subtracting more on the RHS to make it lesser than the LHS
we can clearly see that 3k^2 > 3k, 3k^2 > 1 as k > 1
so we know that what is in the brackets on the RHS is bigger than what is in the brackets on the LHS

hence the result on the RHS must be smaller than the result on the LHS, as we are subtracting more from the k^3 term on the RHS than the LHS

so they pulled it out of thin air just to make the inequality hold? coz if I did this myself I would never ever think of doing that

 

[liamkk112]

so they pulled it out of thin air just to make the inequality hold? coz if I did this myself I would never ever think of doing that

well its not necessarily out of thin air, they chose that specifically because it would make the inequality > 0. it seems random now, but it takes a bit of time to build up intuition on what to do in these inequality questions

there's also not one particular way to do this. for example, we could do this instead:
LHS = 2^(k+1)
= 2 x 2^k
> 2k^3 by assumption

then we can prove the lemma that 2k^3 > (k+1)^3
or that 2k^3 > k^3 + 3k^2 + 3k + 1
or that k^3 > 3k^2 + 3k + 1

let f(k) = k^3 - 3k^2 - 3k - 1
we want to prove that f(k) is increasing for k>=10

differentiating:
f'(k) = 3k^2 - 6k - 3
we want to prove that 3k^2 - 6k - 3 > 0 for k >= 10
(3k - 3)(k - 1) > 0, which is clearly true for k>= 10 by drawing the graph
f'(k) > 0 for k>=10, hence f(k) is increasing for k>= 10
then we can say that f(k) >= f(10) for k >= 10
so k^3 - 3k^2 - 3k - 1 >= 669 > 0 for k >= 10
or that k^3 > 3k^2 + 3k + 1 for k>= 10, which we know means that 2k^3 > (k+1)^3 for k>= 10

hence we see that LHS > 2k^3 > (k+1)^3 = RHS as required, hence by mathematical induction the statement is true for all k >= 10

as you can see there are different methods that require less "intuition" as to what to substitute in the inequality, it just depends if you can see the easy way, or if you want to go through everything i wrote above