Sequence induction proof: u_0 = 3, u_n = (u_{n-1} + 2n^2 - 2)/(n^2) for N*

[Qwertyuiop[]]

A sequence is defined by [imath]u_0 = 3[/imath] and [imath]u_n=\frac{u_{n-1}+2n^2-2}{n^2}[/imath] for N*.
The question is show by induction that for [imath]n \in N, u_n \ge 2.[/imath]. I want to show you my working, i just want to know if it's correct and i didn't make any mistakes in the induction.

Base case: [imath]u_0 = 3[/imath] and [imath]u_0 \ge 2[/imath] so holds for [imath]u_0[/imath].

Inductive hyp: Assuming [imath]u_k \ge 2[/imath], we have to prove that [imath]u_{k+1}\:\ge \:2[/imath].
[imath]u_{k+1}\:is\:given\:by\:u_{k+1}=\:\frac{u_k+2\left(k+1\right)^2-2}{\left(k+1\right)^2}[/imath].

making [imath]u_k[/imath] the subject we get : [imath]u_k\:=\:\left(u_{k+1}\right)\left(k+1\right)^2-2\left(k+1\right)^2+2[/imath] and using the assumption [imath]u_k \ge 2[/imath] ; [imath]\left(u_{k+1}\right)\left(k+1\right)^2-2\left(k+1\right)^2+2 \ge 2 \\ \left(u_{k+1}\right)\left(k+1\right)^2-2\left(k+1\right)^2 \ge 0 \\ \left(k+1\right)^2\:\left(u_{k+1}\:-2\right)\:\ge 0[/imath] . Dividing by [imath]\left(k+1\right)^2[/imath] because it's positive so the sign of inequality is conserved and because we have n >1 we are not dividing by zero.

so [imath]u_{k+1}\ge 2[/imath]

if it's true for [imath]u_{k+1}[/imath] then it must be true [imath]\forall n \in N[/imath].

 

[blamocur]

Don't see any problems with you proof.

 

[Qwertyuiop[]]

Don't see any problems with you proof.

There is just one thing bugging me, the function is defined for N* , we are excluding 0 meaning [imath]n \ge 1[/imath] but they want us to prove for [imath]n \in N[/imath] which would include 0. Was it a typo?

 

[Dr.Peterson]

There is just one thing bugging me, the function is defined for N* , we are excluding 0 meaning [imath]n \ge 1[/imath] but they want us to prove for [imath]n \in N[/imath] which would include 0. Was it a typo?

First, please confirm the definitions of N and N* as you are using them; the terminology and notation can vary. I understand you to be saying that N is the non-negative integers, and N* is the positive integers. Is that correct?

The definition of the function (i.e. sequence) says that the recursion is defined for positive n, which is appropriate, since it doesn't apply to n=0:

A sequence is defined by [imath]u_0 = 3[/imath] and [imath]u_n=\frac{u_{n-1}+2n^2-2}{n^2}[/imath] for N*.
The question is show by induction that for [imath]n \in N, u_n \ge 2.[/imath]. I want to show you my working, i just want to know if it's correct and i didn't make any mistakes in the induction.

(Of course, you meant to say "... for n∈N*. I presume that's your own typo.)

That doesn't mean the sequence is only defined for n∈N*. The initial value together with the recursion covers all non-negative integers n.

They want you to prove that for [imath]n \in N, u_n \ge 2[/imath]; they are including n=0 because the sequence is defined for all non-negative integers.

 

[Qwertyuiop[]]

First, please confirm the definitions of N and N* as you are using them; the terminology and notation can vary. I understand you to be saying that N is the non-negative integers, and N* is the positive integers. Is that correct?

The definition of the function (i.e. sequence) says that the recursion is defined for positive n, which is appropriate, since it doesn't apply to n=0:

(Of course, you meant to say "... for n∈N*. I presume that's your own typo.)

That doesn't mean the sequence is only defined for n∈N*. The initial value together with the recursion covers all non-negative integers n.

They want you to prove that for [imath]n \in N, u_n \ge 2[/imath]; they are including n=0 because the sequence is defined for all non-negative integers.

yes, by N i mean N= [imath][0, \infty)[/imath] and N* = [imath][1, \infty)[/imath].

 

[Dr.Peterson]

yes, by N i mean N= [imath][0, \infty)[/imath] and N* = [imath][1, \infty)[/imath].

Well, no, that is not what you mean. Those intervals are sets of real numbers, not just integers.

If you want to use notation, you must use it correctly.

 

[Qwertyuiop[]]

First, please confirm the definitions of N and N* as you are using them; the terminology and notation can vary. I understand you to be saying that N is the non-negative integers, and N* is the positive integers. Is that correct?

The definition of the function (i.e. sequence) says that the recursion is defined for positive n, which is appropriate, since it doesn't apply to n=0:

(Of course, you meant to say "... for n∈N*. I presume that's your own typo.)

That doesn't mean the sequence is only defined for n∈N*. The initial value together with the recursion covers all non-negative integers n.

They want you to prove that for [imath]n \in N, u_n \ge 2[/imath]; they are including n=0 because the sequence is defined for all non-negative integers.

There is a second part to this question which is to show that the sequence is decreasing. If it's decreasing then [imath]u_{n+1}\:\le u_n[/imath] so [imath]u_{n+1}\:-u_n\:\le 0[/imath]. But i get a massive expression, is there another way to show that a sequence is decreasing/increasing?

 

[blamocur]

There is a second part to this question which is to show that the sequence is decreasing. If it's decreasing then [imath]u_{n+1}\:\le u_n[/imath] so [imath]u_{n+1}\:-u_n\:\le 0[/imath]. But i get a massive expression, is there another way to show that a sequence is decreasing/increasing?

It is decreasing for [imath]n\geq 2[/imath]. Try showing that [imath]u_{n-1} - u_n[/imath] is positive.

 

[Qwertyuiop[]]

It is decreasing for [imath]n\geq 2[/imath]. Try showing that [imath]u_{n-1} - u_n[/imath] is positive.

Well, that's the problem, i get this very big expression so i was wondering if there is an easier way to prove that a sequence is increasing/decreasing. I got stuck at the end. [math]u_{n+1}-u_n \\ \frac{u_n+2\left(n+1\right)^2-2}{\left(n+1\right)^2}-\frac{u_{n-1}+2n^2-2}{n^2}\\ \frac{u_n+2n^2+4n}{\left(n+1\right)^2}-\frac{u_{n-1}+2n^2-2}{n^2}\\ \frac{n^2}{n^2}\cdot \frac{u_n+2n^2+4n}{\left(n+1\right)^2}-\frac{u_{n-1}+2n^2-2}{n^2}\cdot \frac{\left(n+1\right)^2}{\left(n+1\right)^2}\\ \frac{n^2\left(u_n+2n^2+4n\right)-\left(n+1\right)^2\left(u_{n-1}+2n^2-2\right)}{n^2\left(n+1\right)^2}[/math]

 

[Dr.Peterson]

Well, that's the problem, i get this very big expression so i was wondering if there is an easier way to prove that a sequence is increasing/decreasing. I got stuck at the end. [math]u_{n+1}-u_n \\ \frac{u_n+2\left(n+1\right)^2-2}{\left(n+1\right)^2}-\frac{u_{n-1}+2n^2-2}{n^2}\\ \frac{u_n+2n^2+4n}{\left(n+1\right)^2}-\frac{u_{n-1}+2n^2-2}{n^2}\\ \frac{n^2}{n^2}\cdot \frac{u_n+2n^2+4n}{\left(n+1\right)^2}-\frac{u_{n-1}+2n^2-2}{n^2}\cdot \frac{\left(n+1\right)^2}{\left(n+1\right)^2}\\ \frac{n^2\left(u_n+2n^2+4n\right)-\left(n+1\right)^2\left(u_{n-1}+2n^2-2\right)}{n^2\left(n+1\right)^2}[/math]

Don't bring [imath]u_{n-1}[/imath] into it! Leave everything in terms of [imath]u_n[/imath].

 

[Qwertyuiop[]]

Don't bring [imath]u_{n-1}[/imath] into it! Leave everything in terms of [imath]u_n[/imath].

So just use this equation [imath]u_{n+1}=\frac{u_n+2\left(n+1\right)^2-2}{\left(n+1\right)^2}[/imath] and rearrange to get the sign of [imath]u_{n+1}-u_n[/imath] ?

 

[blamocur]

So just use this equation [imath]u_{n+1}=\frac{u_n+2\left(n+1\right)^2-2}{\left(n+1\right)^2}[/imath] and rearrange to get the sign of [imath]u_{n+1}-u_n[/imath] ?

It might not be a good idea to ask guidance for every step along the way. You deserve more self-confidence than that: just try the latest suggestion and see where you get. But if you don't succeed we are here to help again.

 

[Dr.Peterson]

Rather than help you further with the current problem, I'd like to show an alternative approach to the original question in this thread.

A sequence is defined by [imath]u_0 = 3[/imath] and [imath]u_n=\frac{u_{n-1}+2n^2-2}{n^2}[/imath] for N*.
The question is show by induction that for [imath]n \in N, u_n \ge 2.[/imath]. I want to show you my working, i just want to know if it's correct and i didn't make any mistakes in the induction.

Base case: [imath]u_0 = 3[/imath] and [imath]u_0 \ge 2[/imath] so holds for [imath]u_0[/imath].

Inductive hyp: Assuming [imath]u_k \ge 2[/imath], we have to prove that [imath]u_{k+1}\:\ge \:2[/imath].
[imath]u_{k+1}\:is\:given\:by\:u_{k+1}=\:\frac{u_k+2\left(k+1\right)^2-2}{\left(k+1\right)^2}[/imath].

making [imath]u_k[/imath] the subject we get : [imath]u_k\:=\:\left(u_{k+1}\right)\left(k+1\right)^2-2\left(k+1\right)^2+2[/imath] and using the assumption [imath]u_k \ge 2[/imath] ; [imath]\left(u_{k+1}\right)\left(k+1\right)^2-2\left(k+1\right)^2+2 \ge 2 \\ \left(u_{k+1}\right)\left(k+1\right)^2-2\left(k+1\right)^2 \ge 0 \\ \left(k+1\right)^2\:\left(u_{k+1}\:-2\right)\:\ge 0[/imath] . Dividing by [imath]\left(k+1\right)^2[/imath] because it's positive so the sign of inequality is conserved and because we have n >1 we are not dividing by zero.

so [imath]u_{k+1}\ge 2[/imath]

if it's true for [imath]u_{k+1}[/imath] then it must be true [imath]\forall n \in N[/imath].

Base case: [imath]u_0 =3\ge 2[/imath].

Inductive hypothesis: [imath]u_k \ge 2[/imath].
We want to prove that [imath]u_{k+1}\:\ge \:2[/imath].

But [imath]u_{k+1}=\dfrac{u_k+2\left(k+1\right)^2-2}{\left(k+1\right)^2}=\dfrac{u_k-2}{\left(k+1\right)^2}+2\dfrac{\left(k+1\right)^2}{\left(k+1\right)^2}=\dfrac{u_k-2}{\left(k+1\right)^2}+2\ge 2[/imath] because [imath]u_k-2\ge0[/imath].

You can use a similar style for the current work, if you wish. I prefer this style, in part because yours looks too much like solving an inequality, and it is easy to get the implications in the wrong direction.

 

[Qwertyuiop[]]

I showed the working for the first question to my teacher and he said that i should use the assumption [imath]u_n\ge 2[/imath] in the equation [imath]u_{n+1}=\frac{u_n+2\left(n+1\right)^2-2}{\left(n+1\right)^2}[/imath], because [imath]u_n\ge 2[/imath], he said if you substitute 2 for [imath]u_n[/imath] it becomes [imath]u_{n+1}\ge \frac{2-2\left(n+1\right)^2-2}{\left(n+1\right)^2}[/imath](equality becomes an inequality). The 2's cancel and we are left with [imath]u_{n+1}\ge \frac{2\left(n+1\right)^2}{\left(n+1\right)^2}[/imath] which simplifies to [imath]u_{n+1}\ge 2[/imath]. He said when we replace [imath]u_n[/imath] with 2 the equality sign changes to an inequality. I didn't know you could do that in maths. My question is : Can you do that and what is the inequality sign we get after substitution, like it's a greater than or less than inequality.

 

[Dr.Peterson]

I showed the working for the first question to my teacher and he said that i should use the assumption [imath]u_n\ge 2[/imath] in the equation [imath]u_{n+1}=\frac{u_n+2\left(n+1\right)^2-2}{\left(n+1\right)^2}[/imath], because [imath]u_n\ge 2[/imath], he said if you substitute 2 for [imath]u_n[/imath] it becomes [imath]u_{n+1}\ge \frac{2-2\left(n+1\right)^2-2}{\left(n+1\right)^2}[/imath](equality becomes an inequality). The 2's cancel and we are left with [imath]u_{n+1}\ge \frac{2\left(n+1\right)^2}{\left(n+1\right)^2}[/imath] which simplifies to [imath]u_{n+1}\ge 2[/imath]. He said when we replace [imath]u_n[/imath] with 2 the equality sign changes to an inequality. I didn't know you could do that in maths. My question is : Can you do that and what is the inequality sign we get after substitution, like it's a greater than or less than inequality.

That's essentially equivalent to what I did; I just did more processing before using the inequality.

Ultimately, you are using properties of inequality like [math]a>b\implies a+c > b+c[/math] and [math]\text{If }c>0\text{, then }a>b\implies ac > bc,[/math] each of which applies to any of the four inequalities, [imath]>,\ge,<,\le[/imath].

I'd rather not express it in terms of "substitution", because that depends on specific features of the expression into which you are substituting (such as not multiplying by a negative number).

I might say [math]u_{n+1}=\frac{u_n+2\left(n+1\right)^2-2}{\left(n+1\right)^2}\ge=\frac{2+2\left(n+1\right)^2-2}{\left(n+1\right)^2}=\dots[/math]