Induction inquiry (1 Viewer)

[QZP]

I am having trouble consolidating my proof. The question is not one of difficulty, but there is a small aspect I'd like to inquire about, so quickly:

Question: Prove 2^n > n^2 for all n >= 5
*Base case holds*
Suppose n = k true, where integer k>= 5
i.e. 2^k > k^2

Problem starts here
For n = k+1,
2^(k+1) = 2(2^k)
> 2(k^2), from supposition
= k^2 + k^2
> k^2 + 2k +1, since k^2 > 2k + 1
Hence true for n = k+1

For the part underline, do I have to do another proof by induction to show that k^2 > 2k + 1? I know there exists other methods to approach the question (which I have found to avoid the problem) but I'd like to finish this method. Help please

 

[RealiseNothing]

For we get and therefore:

 

[Carrotsticks]

PROTIP: For proving induction inequalities, it's often much easier to consider LHS - RHS and prove it's > 0.

 

[RealiseNothing]

Alternatively:



Since we get

 

[RealiseNothing]

Alternatively 2:

Let

and

increases faster than (just check derivatives for and it's pretty obvious).

 

[RealiseNothing]

Alternatively 3:

For we get









I'm bored, just waiting for carrot to post BOS trial results tbh.

 

[QZP]

You are very good. Thank you
Ah... alternative 2 was intuitive to me though I shouldn't rely on intuition and hence wondered if I had to do another induction proof. But in fact I just had to look at derivatives. + rep

 

[QZP]

How do I integrate this into my proof? Just draw an arrow from "since k^2 > 2k +1" to the side and say "when k =5, k^2 > 2k + 1. Now consider dy/dk [k^2 > 2k + 1] = 2k > 2 hence k^2 > 2k+1 since k>=5"? Is the dy/dk operator even viable in inequalities like that? Ahhhh I'm so bad at math

 

[Sy123]

How do I integrate this into my proof? Just draw an arrow from "since k^2 > 2k +1" to the side and say "when k =5, k^2 > 2k + 1. Now consider dy/dk [k^2 > 2k + 1] = 2k > 2 hence k^2 > 2k+1 since k>=5"? Is the dy/dk operator even viable in inequalities like that? Ahhhh I'm so bad at math

Differentiating does not necessarily hold the inequality

Counter example:

2x + 100 > 5x - 100 (for certain values of x)
but

2 > 5 is not true for any

 

[QZP]

Differentiating does not necessarily hold the inequality

Counter example:

2x + 100 > 5x - 100 (for certain values of x)
but

2 > 5 is not true for any

But it works for the inequality k^2 > 2k + 1 right? If the base case k=5 holds and then you differentiate it i.e
2k>2
k>1
Therefore inequality holds for all k>=5?

 

[RealiseNothing]

But it works for the inequality k^2 > 2k + 1 right? If the base case k=5 holds and then you differentiate it i.e
2k>2
k>1
Therefore inequality holds for all k>=5?

Differentiate each function separately, don't introduce an inequality sign until the very end. That's basically what Sy is saying.

 

[QZP]

Differentiate each function separately, don't introduce an inequality sign until the very end. That's basically what Sy is saying.

Will do. So how do I incorporate that section into my proof? Do I just draw an arrow to the side explaining why or should it be included in the main flow of the proof (but that makes it harder to follow i.e. not clean)?

 

[QZP]

Bump. Need answer to my simple question :S

 

[RealiseNothing]

Will do. So how do I incorporate that section into my proof? Do I just draw an arrow to the side explaining why or should it be included in the main flow of the proof (but that makes it harder to follow i.e. not clean)?

I would personally do it on the side.