HSC Stuff > HSC Mathematics Extension 1

Inequality Induction

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Lottie99:
This is one from fitzpatrick i think?

edmododragon:

--- Quote from: Lottie99 on October 07, 2016, 02:15:43 pm ---This is one from fitzpatrick i think?

--- End quote ---

Interesting, I'm struggling to get this one to work with induction, though I can see an easy way to do it without induction :o

jamonwindeyer:

--- Quote from: edmododragon on October 07, 2016, 02:24:11 pm ---Interesting, I'm struggling to get this one to work with induction, though I can see an easy way to do it without induction :o

--- End quote ---

Yep, this you can do with theory on quadratics, but a solution is coming for induction now! ;D

jamonwindeyer:

--- Quote from: Lottie99 on October 07, 2016, 02:15:43 pm ---This is one from fitzpatrick i think?

--- End quote ---

Cool! So I'll assume you are okay with the first test and setting up the assumption, so I'll take it from:



So you are probably okay up to here right? Substitute \(n=k+1\) and simplify. Remember the aim is to prove that this expression is larger than zero, always, using our assumption. The way we do this is purely down to intuition, and can at times seem a little silly, but what we'll do here is this.

We know (or assume) from our assumption that \(k^2-11k+30\). So let's try and make that appear in the expression:



So, using our induction assumption, we've proven the result for \(n=k+1\), provided that \(k\ge5\). This is a little strange for an induction proof, but what it means is that we test \(k=1,2,3,4\) and \(5\) manually, and then induction takes care of the rest.

This is a little atypical for an induction proof, and I don't see any neater way of using induction to do it. You could do something different with the factorisation approach in the latter steps, but you'd end up just doing the typical quadratic solve anyway:



So that seems redundant, hence why I resorted to this weird method. It's more, induction-y ;) This isn't the best example of an induction question imo, I don't blame you for being confused by it, got another? :)

edmododragon:

--- Quote from: jamonwindeyer on October 07, 2016, 02:48:28 pm ---Cool! So I'll assume you are okay with the first test and setting up the assumption, so I'll take it from:



So you are probably okay up to here right? Substitute \(n=k+1\) and simplify. Remember the aim is to prove that this expression is larger than zero, always, using our assumption. The way we do this is purely down to intuition, and can at times seem a little silly, but what we'll do here is this.

We know (or assume) from our assumption that \(k^2-11k+30\). So let's try and make that appear in the expression:



So, using our induction assumption, we've proven the result for \(n=k+1\), provided that \(k\ge5\). This is a little strange for an induction proof, but what it means is that we test \(k=1,2,3,4\) and \(5\) manually, and then induction takes care of the rest.

This is a little atypical for an induction proof, and I don't see any neater way of using induction to do it. You could do something different with the factorisation approach in the latter steps, but you'd end up just doing the typical quadratic solve anyway:



So that seems redundant, hence why I resorted to this weird method. It's more, induction-y ;) This isn't the best example of an induction question imo, I don't blame you for being confused by it, got another? :)

--- End quote ---

That quadratic proof is what I was thinking of. Do you think they would ever ask a question in the HSC where you would need to test more than the initial case (excluding the inductions in Ext2 where each value is dependent on the previous two)?

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