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Let the sequence $G_0, G _1, G_2, ...$ be defined recursively as follows:

$G_0=0, G_1=1,$ and $G_n=5G_{n-1}-6G_{n-2}$ for every $n\in\mathbb{N}, n\ge2$.

Prove that for all $n\in\mathbb{N}$, $G_n=3^n-2^n$.

Proof. By strong induction. Let the induction hypothesis, $P(n)$, be $G_n=3^n-2^n$

Base Case: For $(n=0)$, $P(0)$ is true because $3^0-2^0 =0$

For $(n=1)$, $P(1)$ is true because $3^1-2^1=1$

Inductive Step: Assume that $P(n-1)$ and $P(n-2)$, where $n\ge2$, are true for purposes of induction.

So, we assume that $G_{n-1}=3^{n-1}-2^{n-1}$ and $G_{n-2}=3^{n-2}-2^{n-2}$, and we must show that $G_{ n }=3^{ n }-2^{ n }$.

Since we assumed $P(n-1)$ and $P(n-2)$, we can rewrite $G_n=5G_{n-1}-6G_{n-2}$ as $G_n=5(3^{n-1}-2^{n-1})-{ 6 }(3^{n-2}-2^{n-2})$

So, we get:

$\Rightarrow G_n=5\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }-(\frac { 6 }{ 3 } \cdot 3^{ n-1 }-\frac { 6 }{ 2 } \cdot 2^{ n-1 })$

$\Rightarrow G_n=5\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }-2\cdot 3^{ n-1 }+3\cdot 2^{ n-1 }$

$\Rightarrow G_n=5\cdot 3^{ n-1 }-2\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }+3\cdot 2^{ n-1}$

$\Rightarrow G_n=3\cdot 3^{ n-1 }-2\cdot 2^{ n-1 }$

$\Rightarrow G_n=\frac { 1 }{ 3 } \cdot 3\cdot 3^n-\frac { 1 }{ 2 } \cdot 2\cdot 2^n$

$\Rightarrow G_n=3^n-2^n$

The only real issue I have at this point is that I don't know how to properly conclude this proof with a final statement. A hint/guidance in that regard would be much appreciated.

In addition, please feel free to offer advice and/or constructive criticism about my proof.

Cherry_Developer
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    It looks great, you have effectly showed that if the property $G_{i}=3^i-2^i$ holds for $i\in{1,2,\dots n-1}$ then $G_{n}=3^n-2^n$ also. – Asinomás Aug 03 '16 at 17:12
  • How about QED. Or "as desired" or a "tombstone." Your proof is great. Small note, "strong induction" means that we are assuming that $P(k)$ is true for all $k<n$ whereas standard induction, we only are assuming it to be true for as many cases as necessary. – Doug M Aug 03 '16 at 17:14
  • @DougM Does this mean that this was a proof by ordinary induction? – Cherry_Developer Aug 03 '16 at 17:19
  • Correct, but, from a logical point of view, the distinction is trivial. The reason for the distinction is so that the reader of the proof is absolutely clear on what has been assumed at the inductive step. – Doug M Aug 03 '16 at 17:25
  • @DougM I am new to proofs by induction, so please correct me if I'm wrong.

    I am assuming that $P(0), P(1), ..., P(n-2), P(n-1)$ together imply $P(n)$. Does that not adhere to the Principle of Strong Induction? Or am I mixing something up here?

     – Cherry_Developer Aug 03 '16 at 17:28
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    In your description of $P(n)$, the part "for all $n$ $\dots$" is at best superfluous, and at worst confusing or incorrect. Better, for any integer $k\ge 0$ let $P(k)$ be the assertion $G(k)=3^k-2^k$. – André Nicolas Aug 03 '16 at 17:32
  • @Cherry_Developer It is just slightly overkill, you only need to assume $P(n-1),P(n-2)$ are true to prove $P(n).$ – Doug M Aug 03 '16 at 17:35
  • You need to assume $,n\ge2,$ else $,P(n-2),$ is not defined. See this question for the type of errors this can cause. – Bill Dubuque Aug 03 '16 at 17:37
  • @AndréNicolas I don't understand why the introduction of $k$ is necessary. – Cherry_Developer Aug 03 '16 at 17:42
  • @BillDubuque Is it sufficient to add it into the first line of the Inductive Step as such:

    Inductive Step: Assume that $P(n−1)$ and $P(n−2)$, where $n\ge2$, are true for purposes of induction.

     – Cherry_Developer Aug 03 '16 at 17:44
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    Not necessary. But the deletion of "for all $n$ $\dots$" is. – André Nicolas Aug 03 '16 at 17:47
  • @Cherry That is one way to fix the problem. Is this use of "base case" in strong induction your own wording or did you read that somewhere? – Bill Dubuque Aug 03 '16 at 17:48
  • @BillDubuque I read it in the Mathematics for Computer Science course notes on MITOCW. You can find it in section 3.4 (starts on page 64). This problem is from one of the course assignments too. I am attempting to abide by what the course notes and lectures profess since I do not have a professor or TA to guide me. – Cherry_Developer Aug 03 '16 at 17:55
  • @Cherry_Developer Standard or "weak" induction in this case means you only need $P(n-1)$ and $P(n-2)$ to be true in order to show that $P(n)$ is true. It is not necessary to involve $P(n-3), P(n-4), P(n-5), \ldots$ directly in your calculations. To better see the difference, try to prove "Every integer greater than $1$ is divisible by a prime." – Marcus Andrews Aug 03 '16 at 17:56
  • @AndréNicolas I want to understand why it is wrong. Is the $\forall n$ contradictory to the rest of the description of $P(n)$? Namely, the fact that $n$ has to be in the natural numbers and greater than or equal to $0$? Or is it something else? – Cherry_Developer Aug 03 '16 at 18:04
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    @Cherry Thanks for the link. I see that they state strong induction with a base case P(0). But often we don't need base case(s) for strong induction and the induction principle can be stated without any. For example, every integer > 1 is a product of primes. Suppose for induction it is true for all naturals < n. If n is prime we are done, else n is composite so it is a product of smaller naturals n = ab so by induction a,b are products of primes. Appending their products shows that n is a product of primes. No base case! – Bill Dubuque Aug 03 '16 at 18:05
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    The assertion $P(n)$ is not the assertion that $G(n)=3^n-2^n$ for all $n$. If we write the latter in symbols, it is $\forall n(G(n)=3^n-2^n)$. Now $n$ is a "dummy variable" which gets quantified out. We want that for any particular $n$, $P(n)$ is the assertion $\dots$. – André Nicolas Aug 03 '16 at 18:12
  • @BillDubuque If you take a look at the notes again, on the bottom of page 65, there is a proof by strong induction of the same theorem that you just mentioned to me that utilized the base case of $P(2)$. Is that unnecessary?

    I've been getting somewhat ambiguous responses regarding whether or not the proof I wrote up actually required Strong Induction. Would you be able to clarify this for me?

     – Cherry_Developer Aug 03 '16 at 18:28
  • @AndréNicolas I have edited my original write-up. I think I now understand what my mistake was. Thank you for your help. – Cherry_Developer Aug 03 '16 at 18:38
  • @Cherry Yes, I saw that. They have to do it that way because they have to conform to the strong induction rule they gave with a base case. But as you can see the handling of the base case is redundant since the general argument in the inductive step already handles the case when n is prime. Essentially all primes are base cases. The proof works for any property that is multiplicative, ie.. $P(a),P(b),\Rightarrow, P(ab).,$ It lifts the property from the generators (primes) to all products (it is structural induction in a monoid = set closed under multiplication). – Bill Dubuque Aug 03 '16 at 18:41
  • Your induction hypothesis should be that P(k) is true for all k < n. If you use the form in the notes then you need to prove the base case P(0) and then do the case P(1) as a special case in the inductive step, before using the recursion (since it requires two prior values, which you don't have when n = 1). – Bill Dubuque Aug 03 '16 at 18:53
  • @BillDubuque So, in accordance with that paragraph underneath the Principle of Strong Induction on page 65 of the notes I am following, I am indeed using strong induction, yes? I am assuming $P(n-2)$ and $P(n-1)$ are both true in order to prove $P(n)$. Or can one use ordinary induction in this particular case? This is what is really confusing me. – Cherry_Developer Aug 03 '16 at 19:32
  • @BillDubuque I appreciate all of the help and guidance you have so generously provided me with. If you do not mind, please post some (or all) of the things you have already written in these comments in an answer below. I will be sure to accept it and +1. It's the least I could do. Thank you. – Cherry_Developer Aug 03 '16 at 19:40

2 Answers2

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You have the right idea, but there are some minor points that need correction.

The strong induction principle in your notes is stated as follows:

Principle of Strong Induction $\ $ Let $\,P(n)\,$ be a predicate. If

  • $\ P(0)$ is true, and

  • for all $\,n\in \Bbb N,\ P(0), P(1),\ldots, P(n)\,$ together imply $\,P(n\!+\!1)\,$ then $\,P(n)\,$ is true for all $\,n\in\Bbb N$

Your $\,P(n)\,$ is $\, G_n = 3^n - 2^n.\,$ You have verified that $\,P(0)\,$ is true.

Your induction hypothesis is that $\,P(k)\,$ is true for all $k \le n.\,$ You have essentially shown that $\,P(n\!-\!1),P(n)\,\Rightarrow\,P(n\!+\!1)\,$ but that only works for $\,n\ge 1$ (else $\,P(n-1)\,$ is undefined). Thus you need to separately verify $\,P(1)\,$ (to be pedantic, this is part of the inductive step, not the base case, according to the above formulation of strong induction, though that is a somewhat arbitrary distinction)

It is illuminating to observe that the recurrence in the induction is a special case of

$$ a^{n+1}-b^{n+1} =\, (a+b)(a^n-b^n) -ab (a^{n-1} - b^{n-1})$$

which can be verified directly or derived from the fact that $\,a,b\,$ are roots of

$$(x\!-\!a)(x\!-\!b) = x^2\! - (a\!+\!b) x + ab\,\Rightarrow\, x^{n+1}\! = (a\!+\!b)\,x^n - ab\, x^{n-1}$$

The proof will be simpler (and more insightful) if you work with this general case, i.e. prove that $\,f_n = a^n - b^n\,$ satisfies $\,f_{n+1} = (a+b) f_{n} - ab f_{n-1},\ f_0 = 1,\ f_1 = a-b\,$ for all $\,n\ge 0.\,$ Then your problem is just the special case $\,a,b = 3,2,\,$ and the inductive step is much clearer.

Bill Dubuque
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  • I was under the impression that I showed that $P(n-2),P(n-1)\Rightarrow P(n)$. You stated that I showed $P(n-1),P(n)\Rightarrow P(n+1)$. Why? – Cherry_Developer Aug 03 '16 at 19:57
  • @Cherry That's why I said "essentially". Substitute $,n+1,$ for $,n,$ in your proof to get the upshifted form. I wrote the induction in the above form used in the MIT notes. – Bill Dubuque Aug 03 '16 at 20:01
  • Ah ok. I apologize for the slew of questions. I would just much rather struggle with the math, and ask these questions now, than when I actually take discrete math. Thank you for all your help. – Cherry_Developer Aug 03 '16 at 20:07
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    @Cherry_Developer It's the nature of the beast to struggle with induction proofs when one first encounters them (evolution doesn't program our minds for such). Many fit into particular patterns that are easier to comprehend in the abstract (such as the above which is essentially exploiting the uniqueness theorem for recurrences).. Another common form of induction is telescopy, e.g. see here for a vivid 2D example. – Bill Dubuque Aug 03 '16 at 20:11
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Yes, your proof is perfectly fine. Good job! You can write something like "The assertion follows.". But honestly it isn't necessary since it is in this case pretty simple for readers to see where the proof is complete (after the inductive step).

Yaddle
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