A Question about Defining a Sequence "Inductively"

[The Student]

Question: Consider the Fibonacci sequence defined inductively by a(0) = 1, a(1) = 1, and a(n+1) = a(n) + a(n - 1) for n ≥ 1. Consider the ratio of consecutive terms, b(n) = a(n+1)/a(n), for n ≥ 0. Since n ≥ 1,
b(n) = 1 + 1/b(n-1), show that 1 ≤ b(n) ≤ 2 for all n ≥ 0. The answer: The fact that b(0) = 1/1 = 1 we see that b(n) ≥ 1 for all n ≥ 0. Thus b(n - 1) ≥ 1 for all n ≥ 1. This implies that b(n) ≤ 1 + 1/1 = 2 ∀n≥ 1, so 1 ≤ b(n) ≤ 2 for all n ≥0.
My problem is where the answer claims that "b(n) ≥ 1 for all n ≥ 0". The only way that this seems possible is because of the description "defined inductively"put in the beginning of the question? But I don't know what that really means.

 

[stapel]

Consider the ratio of consecutive terms, b(n) = a(n+1)/a(n), for n ≥ 0. Since n ≥ 1, b(n) = 1 + 1/b(n-1)....

Verify their alternative form for b(n). You are given the definition:

. . . . .\(\displaystyle b_n\, =\, \dfrac{a_{n+1}}{a_n}\)

Show that this is indeed equivalent to:

. . . . .\(\displaystyle b_n\, =\, 1\, +\, \dfrac{1}{b_{n-1}}\)

Since the terms b(n) are all positive (because they're ratios of positive values), and because "1 plus (a positive value)" always has to be greater than "just 1", then the bit you're questioning is proven from the definition.

 

[The Student]

Verify their alternative form for b(n). You are given the definition:

. . . . .\(\displaystyle b_n\, =\, \dfrac{a_{n+1}}{a_n}\)

Show that this is indeed equivalent to:

. . . . .\(\displaystyle b_n\, =\, 1\, +\, \dfrac{1}{b_{n-1}}\)

Since the terms b(n) are all positive (because they're ratios of positive values), and because "1 plus (a positive value)" always has to be greater than "just 1", then the bit you're questioning is proven from the definition.

But how do we know that a(n+1)/a(n) is positive? I realize that it is given that it starts off positive because a(0) = 1 and a(1) = 1, but how do I know that a(n+1) and a(n) are both positive?

 

[JeffM]

But how do we know that a(n+1)/a(n) is positive? I realize that it is given that it starts off positive because a(0) = 1 and a(1) = 1, but how do I know that a(n+1) and a(n) are both positive?

You can use mathematical induction to prove that a(n + 1)/a(n) > 0

 

[The Student]

You can use mathematical induction to prove that a(n + 1)/a(n) > 0

The part of the answer that says, "we see that b(n) ≥ 1 for all n ≥ 0" seems to expect that we already know this from something said in the question. Do you think that it might have something to do with the term, "defined inductively" in the beginning of the question?

 

[pka]

But how do we know that a(n+1)/a(n) is positive? I realize that it is given that it starts off positive because a(0) = 1 and a(1) = 1, but how do I know that a(n+1) and a(n) are both positive?

\(\displaystyle a_0=1~ \&~ a_1=1\) and if \(\displaystyle n\ge 1\) then \(\displaystyle a_{n+1}=a_n+a_{n-1}\).

This is also known as a recessive definition. It is used extensively is programming.

Now to your question: if we add to non-negative integers then you get a positive integer.

Thus if \(\displaystyle n\ge 1\) you get \(\displaystyle 0<a_n<a_{n+1}\).

Surely you can see \(\displaystyle n\ge 1\) then \(\displaystyle b_n\ge 1~.\)

 

[Deleted member 4993]

The part of the answer that says, "we see that b(n) ≥ 1 for all n ≥ 0" seems to expect that we already know this from something said in the question. Do you think that it might have something to do with the term, "defined inductively" in the beginning of the question?

It may not be mathematically rigorous - but I would argue this way:

1) Suppose a_m+1 < a_m - for the first time - before that we had a_k+1 > a_k where k < m

2) Since a_m+1 = a_m + a_m-1 → a_m-1 < 0 for the first time

3) But a_m-1 = a_m-2 + a_m-3

4) a_m-2 & a_m-3 are positive numbers (a_m-1 first assumed negative number of the sequence) and their sum cannot be negative

5) Reductio ad absurdum

so we will have a_m+1 > a_m → a_m+1/ a_m > 1→ b_m = a_m+1/ a_m > 1

 

[The Student]

\(\displaystyle a_0=1~ \&~ a_1=1\) and if \(\displaystyle n\ge 1\) then \(\displaystyle a_{n+1}=a_n+a_{n-1}\).

This is also known as a recessive definition. It is used extensively is programming.

Now to your question: if we add to non-negative integers then you get a positive integer.

How do you know that we will be adding only positive numbers to a(0) and a(1)?

Thus if \(\displaystyle n\ge 1\) you get \(\displaystyle 0<a_n<a_{n+1}\).

Surely you can see \(\displaystyle n\ge 1\) then \(\displaystyle b_n\ge 1~.\)

How can we know that a(n+1) or a(n) is positive for all n? The answer seems to expect that we know this from the question. I am starting to think that it has something to do with the description, "defined inductively".

If you remember the last time that I was on this forum, I had a similar problem. I never did know how a(n+1) = (a(n) + 2)^(1/2) is positive, but the same term came up which is "defined inductively".

 

[daon2]

How do you know that we will be adding only positive numbers to a(0) and a(1)?



How can we know that a(n+1) or a(n) is positive for all n? The answer seems to expect that we know this from the question. I am starting to think that it has something to do with the description, "defined inductively".

If you remember the last time that I was on this forum, I had a similar problem. I never did know how a(n+1) = (a(n) + 2)^(1/2) is positive, but the same term came up which is "defined inductively".

It might benefit you to show a(n) is a linear combination of a(0) and a(1) (That is technically the point.)

a(2) = 1*a(1)+1*a(0)
a(3) = a(2)+a(1) = 2*a(1)+1*a(0)
a(4) = a(3)+a(2) = 3*a(1)+2*a(0)


etc

 

[The Student]

It might benefit you to show a(n) is a linear combination of a(0) and a(1) (That is technically the point.)

a(2) = 1*a(1)+1*a(0)
a(3) = a(2)+a(1) = 2*a(1)+1*a(0)
a(4) = a(3)+a(2) = 3*a(1)+2*a(0)

etc

Normally, I can accept a conclusion when it is backed by a rule or property that I can identify as necessary to produce the result.

So my issue is that we don't know what a(n) is, so how do we know that it won't start becoming negative at some point?

The answer does indicate that we "know" this, but it is not clear to me how. Do you think that it has something to do with the description, "defined inductively"?

 

[daon2]

Normally, I can accept a conclusion when it is backed by a rule or property that I can identify as necessary to produce the result.

So my issue is that we don't know what a(n) is, so how do we know that it won't start becoming negative at some point?

The answer does indicate that we "know" this, but it is not clear to me how. Do you think that it has something to do with the description, "defined inductively"?

No, it is just how it is defined. a(0) and a(1) are positive integers. All a(n) are defined as the sum of the previous 2 positive integers in the list. You don't need to know what a(n) is.

Maybe it isn't "obvious" if pure rigor is needed to convince you, and a proof should be shown that a(n) is positive for all n. The proof is: a(0) >0, a(1) > 0 so a(2)=a(1)+a(2)>0. Now assume a(k) is nonzero for all k=1,2,...,n. Then a(n+1)=a(n)+a(n-1)>0+0=0. It's that simple.

 

[The Student]

Thanks everyone, I think that I get it now. :grin: