Sequence and series

[Kyan]

Hello i need help please,
u0= 1
un+1=( 3un + 2vn)/5
v0 = 2
vn+1 = (2un+3vn)/5
1.Calculate u1,u2,v1,v2
2.We consider the sequence (dn) defined for any natural number n by dn = vn-un
a. Show that the sequence (dn) is a geometric sequence of which we will give its common ratio and its first term.
b. Deduce the expression of dn depending on n
3. We consider the sequence (sn) Defined for any natural number n by sn = un+ vn
a. Calculate s0, s1and s2 . What can we guess?
b. Show that, for all nEN, sn+1= sn. What can we deduce?
4. Deduce an expression of un and vn depending on n.
5.determine depending on n E N
a. Tn= u0 + u1 + ... + un
b. Wn= v0 + v1 + ... + vn

actually i answered the first question and the second and i found the common ratio is 1/5 and the first term 1 so dn= 1*1/5^n
For the 3. I found 3
And i cant resolve the 4 and 5 can you help me

 

[lev888]

Please repost using the subscript button (x₂). un+1 should look like u_n+1

 

[JeffM]

What did you get for an answer to part 1?

 

[HallsofIvy]

Hello i need help please,
u0= 1
un+1=( 3un + 2vn)/5
v0 = 2
vn+1 = (2un+3vn)/5

\(\displaystyle u_0= 1\)
\(\displaystyle u_{n+1}= (3u_n+ 2v_n)/5\)
\(\displaystyle v_0= 2\)
\(\displaystyle v_n= (2u_n+ 3v_n)/5\)

1.Calculate u1,u2,v1,v2[/quote[
\(\displaystyle u_1= (3u_0+ 2v_0)/5= (3(1)+ 2(2))/5= 7/5\)
\(\displaystyle v_1= (2u_0+ 3v_0)/5= (2(1)+ 3(2))/5= 8/5\)
\(\displaystyle u_2= (3u_1+ 2v_1)/5= (3(7/5)+ 2(8/5))/5= (37/5)/5= 37/25\)
\(\displaystyle v_2= (2u_1+ 3v_1)/5= (2(7/5)+ 3(8/5))/5= (38/5)/5= 38/25\)/
That's what you got, right? Just arithmetic.

[/tex]2.We consider the sequence (dn) defined for any natural number n by dn = vn-un
a. Show that the sequence (dn) is a geometric sequence of which we will give its common ratio and its first term.

Well, \(\displaystyle d_0= u_0- v_0= 1- 2= -1\)
\(\displaystyle d_1= u_1- v_1= 7/5- 8/5= -1/5\)
so if it IS a geometric sequence, its first term is -1 and its common ration is 1/5.
Check: \(\displaystyle u_2- v_2= 37/25- 38/25= 1/25= (-1/5)(1/5)\).

Looks good but we need to check in general. We can do that with "proof by induction".

b. Deduce the expression of dn depending on n

Assuming we are correct that \(\displaystyle d_n\) is a geometric sequence with first term -1 and common factor 1/5 then the nth term is \(\displaystyle d_n= -(1/5)^n\).

3. We consider the sequence (sn) Defined for any natural number n by sn = un+ vn
a. Calculate s0, s1and s2 . What can we guess?

So \(\displaystyle s_0= u_0+ v_0= 1+ 2= 3\)
\(\displaystyle s_1= u_1+ v_1= 7/5+ 8/5= 15/5= 3\)
\(\displaystyle s_2= u_2+ v_2= 37/25+ 38/25= 75/25= 3\)
Gosh, what do you guess \(\displaystyle s_n\) will be?

b. Show that, for all nEN, sn+1= sn. What can we deduce?
4. Deduce an expression of un and vn depending on n.

If it is true that \(\displaystyle d_n= u_n- v_n= -(1/5)^n\) and \(\displaystyle s_n= u_n+v_n= 3\)
the \(\displaystyle u_n= (1/2)(dn+ s_n)= (1/2)(-(1/5)^n+ 3)\) and \(\displaystyle v_n= (1/2)(s_n- d_n)= (1/2)(3+ (1/5)^n\).
Check: When n= 0, \(\displaystyle u_0= (1/2)(-1+ 3)= 1\) and \(\displaystyle v_0= (1/2)(3+ 1)= 2\).
When n= 1, \(\displaystyle u_1= (1/2)(-1/5+ 3)= (1/2)(14/5)= 7/5\) and \(\displaystyle v_1= (1/2)(3+ 1/5)= (1/2)(16/5)= 8/5\).
When n= 2, \(\displaystyle u_2= (1/2)(-1/25+ 3)= (1/2)(74/25)= 37/25\) and \(\displaystyle v_2= (1/2)(3+ 1/25)= (1/2)(76/25)= 38/25\). Yes, that is what we had before.

5.determine depending on n E N
a. Tn= u0 + u1 + ... + un

Well, it is \(\displaystyle T_n= (1+ 7/5+ 37/25+ ,,,+ (1/2)(-1/5^n+ 3)= (1/2)(-1- 1/10- 1/25- .,,- 1/5^n)+ (1/2)(3+ 3+ ...+ 3)\). The first of those is a geometric series and the second is 3n.

b. Wn= v0 + v1 + ... + vn

I'll leave this for you. It is done the same way as a.

actually i answered the first question and the second and i found the common ratio is 1/5 and the first term 1 so dn= 1*1/5^n
For the 3. I found 3

Yes, those are correct.

And i cant resolve the 4 and 5 can you help me

 

[Kyan]

H

\(\displaystyle u_0= 1\)
\(\displaystyle u_{n+1}= (3u_n+ 2v_n)/5\)
\(\displaystyle v_0= 2\)
\(\displaystyle v_n= (2u_n+ 3v_n)/5\)

Well, \(\displaystyle d_0= u_0- v_0= 1- 2= -1\)
\(\displaystyle d_1= u_1- v_1= 7/5- 8/5= -1/5\)
so if it IS a geometric sequence, its first term is -1 and its common ration is 1/5.
Check: \(\displaystyle u_2- v_2= 37/25- 38/25= 1/25= (-1/5)(1/5)\).

Looks good but we need to check in general. We can do that with "proof by induction".

Assuming we are correct that \(\displaystyle d_n\) is a geometric sequence with first term -1 and common factor 1/5 then the nth term is \(\displaystyle d_n= -(1/5)^n\).


So \(\displaystyle s_0= u_0+ v_0= 1+ 2= 3\)
\(\displaystyle s_1= u_1+ v_1= 7/5+ 8/5= 15/5= 3\)
\(\displaystyle s_2= u_2+ v_2= 37/25+ 38/25= 75/25= 3\)
Gosh, what do you guess \(\displaystyle s_n\) will be?


If it is true that \(\displaystyle d_n= u_n- v_n= -(1/5)^n\) and \(\displaystyle s_n= u_n+v_n= 3\)
the \(\displaystyle u_n= (1/2)(dn+ s_n)= (1/2)(-(1/5)^n+ 3)\) and \(\displaystyle v_n= (1/2)(s_n- d_n)= (1/2)(3+ (1/5)^n\).
Check: When n= 0, \(\displaystyle u_0= (1/2)(-1+ 3)= 1\) and \(\displaystyle v_0= (1/2)(3+ 1)= 2\).
When n= 1, \(\displaystyle u_1= (1/2)(-1/5+ 3)= (1/2)(14/5)= 7/5\) and \(\displaystyle v_1= (1/2)(3+ 1/5)= (1/2)(16/5)= 8/5\).
When n= 2, \(\displaystyle u_2= (1/2)(-1/25+ 3)= (1/2)(74/25)= 37/25\) and \(\displaystyle v_2= (1/2)(3+ 1/25)= (1/2)(76/25)= 38/25\). Yes, that is what we had before.


Well, it is \(\displaystyle T_n= (1+ 7/5+ 37/25+ ,,,+ (1/2)(-1/5^n+ 3)= (1/2)(-1- 1/10- 1/25- .,,- 1/5^n)+ (1/2)(3+ 3+ ...+ 3)\). The first of those is a geometric series and the second is 3n.


I'll leave this for you. It is done the same way as a.


Yes, those are correct.

hello but i don’t understand why did you do dn= un-vn when actually dn= vn-un and i got 1*(1/5)^n

 

[Kyan]

What did you get for an answer to part 1?

I got u₁= 7/5
u₂= 37/25
v₁= 8/5
v₂= 38/25

 

[JeffM]

I got u₁= 7/5
u₂= 37/25
v₁= 8/5
v₂= 38/25

[MATH]u_0 = 1 \text { and } v_0 = 2.[/MATH]
[MATH]n > 1 \implies u_{n+1} = \dfrac{3u_n+ 2v_n}{5} \text { and } v_{n+1} = \dfrac{2u_n + 3v_n}{5}.[/MATH]
So we can conclude that

[MATH]u_1 = \dfrac{3 * 5 / 5 + 2 * 10 / 5}{5} = 7/5, \ v_1 = \dfrac{2 * 5 / 5 + 3 * 10 / 5}{5} = 8/5,[/MATH]
[MATH]u_2 = \dfrac{3 * 7 / 5 + 2 * 8 / 5}{5} = 37/25, \text { and } v_2 = \dfrac{2 * 7 / 5 + 3 * 8 / 5}{5} = 38/25.[/MATH]
So you were fine on part 1. Well done.

Then on part 2, you presumably went

[MATH]d_n = v_n - u_n \implies d_0 = -1,\ d_1 = -\dfrac{1}{5}, \text { and } d_2 = \dfrac{1}{25}.[/MATH]
You then "saw" that you were dealing with a geometric sequence with a common ratio of 1/5 and an initial term of 1. Nice. If your book defines the natural numbers as starting with zero, then your formula of

[MATH]d_n = - \left ( \dfrac{1}{5} \right )^n[/MATH] is also correct.

Seeing is not proving. I am not sure whether your book wants proof. "Deduce" sounds like "prove" to me.

Part 3 is even easier, and again you are correct in what you have posited, but it is not proof.

Part 5 depends on part 4 so we must deal with that before proceeding.

We know the first three terms of the u sequence. Do we get any hints from those? As an initial experiment, let's take first differences. This may require us to calculate at least one more term in the sequence. Let's do that for u₃.

[MATH]u_3 = \dfrac{3 * 37/25 + 2 * 38/25}{5} = 187/125.[/MATH]
First differences
7/5 - 1 = 2/5, 37/25 - 7/5 = 37/25 - 35/25 = 2/25, and 187/125 - 185/125 = 2/125.

What if anything does that suggest? Should we take second differences?

 

[Kyan]

[MATH]u_0 = 1 \text { and } v_0 = 2.[/MATH]
[MATH]n > 1 \implies u_{n+1} = \dfrac{3u_n+ 2v_n}{5} \text { and } v_{n+1} = \dfrac{2u_n + 3v_n}{5}.[/MATH]
So we can conclude that

[MATH]u_1 = \dfrac{3 * 5 / 5 + 2 * 10 / 5}{5} = 7/5, \ v_1 = \dfrac{2 * 5 / 5 + 3 * 10 / 5}{5} = 8/5,[/MATH]
[MATH]u_2 = \dfrac{3 * 7 / 5 + 2 * 8 / 5}{5} = 37/25, \text { and } v_2 = \dfrac{2 * 7 / 5 + 3 * 8 / 5}{5} = 38/25.[/MATH]
So you were fine on part 1. Well done.

Then on part 2, you presumably went

[MATH]d_n = v_n - u_n \implies d_0 = -1,\ d_1 = -\dfrac{1}{5}, \text { and } d_2 = \dfrac{1}{25}.[/MATH]
You then "saw" that you were dealing with a geometric sequence with a common ratio of 1/5 and an initial term of 1. Nice. If your book defines the natural numbers as starting with zero, then your formula of

[MATH]d_n = - \left ( \dfrac{1}{5} \right )^n[/MATH] is also correct.

Seeing is not proving. I am not sure whether your book wants proof. "Deduce" sounds like "prove" to me.

Part 3 is even easier, and again you are correct in what you have posited, but it is not proof.

Part 5 depends on part 4 so we must deal with that before proceeding.

We know the first three terms of the u sequence. Do we get any hints from those? As an initial experiment, let's take first differences. This may require us to calculate at least one more term in the sequence. Let's do that for u₃.

[MATH]u_3 = \dfrac{3 * 37/25 + 2 * 38/25}{5} = 187/125.[/MATH]
First differences
7/5 - 1 = 2/5, 37/25 - 7/5 = 37/25 - 35/25 = 2/25, and 187/125 - 185/125 = 2/125.

What if anything does that suggest? Should we take second differences?

i didn’t understand why you made the difference and why for d₀= -1you have a negative number when v_n is bigger than u_n

 

[JeffM]

i didn’t understand why you made the difference and why for d₀= -1you have a negative number when v_n is bigger than u_n

That d₀ was a typo. I first thought that we were dealing with u_n - v_n. When I noticed that I had it backwards, I had to change signs, but obviously forgot to do so on d₀. Sorry about that.

Are you asking what I mean about first differences with respect to a sequence?

Do you see that having done that, an obvious pattern emerged?

 

[Kyan]

---

That d₀ was a typo. I first thought that we were dealing with u_n - v_n. When I noticed that I had it backwards, I had to change signs, but obviously forgot to do so on d₀. Sorry about that.

Are you asking what I mean about first differences with respect to a sequence?

Do you see that having done that, an obvious pattern emerged?

Yes, but I’m sorry i don’t see the obvious pattern

 

[JeffM]

First, I said taking first differences was an experiment. It was something to try. But it sometimes works. (Sometimes taking second differences, the differences between the first differences, works: you do not know until you have tried.)

Second, what did these initial experiments show

[MATH]u_1 - u_0 = \dfrac{2}{5}[/MATH]
[MATH]u_2 - u_1 = \dfrac{2}{25}[/MATH]
[MATH]u_3 - u_2 = \dfrac{2}{125}[/MATH]
You literally do not see a pattern there? Now I do not know exactly how to use that pattern yet, but it sure looks like

[MATH]u_{n+1} - u_n = \left ( \dfrac{2^{\{1/(n+1)\}}}{5} \right )^{(n+1)}[/MATH] describes the first differences.

That is speculation, not proof, but it looks like a pretty strong hint. The difficulty then becomes how to use the pattern on the u's rather than the first differences. Because the v's and the u's are defined in terms of each other, I would see what first differences tell us about the v's. And then we have further clues in the d's and s's. You have to try things; inspiration does not come from staring at a blank sheet of paper. Tell me what you learn about the first differences of the v's.