Calculating $U_{513}$ of a Sequence Defined by Recurrence Relation

[anemone]

Let $U_1,\,U_2,\,\cdots$ be a sequence defined by $U_1=1$ and for $n>1$, $U_{n+1}=\sqrt{U_n^2-2U_n+3}+1$. Find $U_{513}$.

 

[topsquark]

Well it's 33 isn't it? The solution is so obvious that I just can't bear to post it. (Angel)

Where do you come up with these things??

-Dan

 

[mathbalarka]

Well it's 33 isn't it?

Close, but no cigar. $U_{510}$ is 33, not $U_{513}$, as 510 is 2 modulo 4, unlike 513.

(Lipssealed) I won't give out anything that will lead to a solution.

EDIT : This is flawed too. :(

 

[topsquark]

Close, but no cigar. $U_{510}$ is 33, not $U_{513}$, as 510 is 2 modulo 4, unlike 513.

(Lipssealed) I won't give out anything that will lead to a solution.

Hmph. I got my count wrong. No more Excel for me!

-Dan

 

[MarkFL]

My solution:

Spoiler

If we square the recursion, we may write:

$$\left(U_{n+1}-1\right)^2=\left(U_{n}-1\right)^2+2$$

Let:

$$V_n=\left(U_{n}-1\right)^2$$

And there results:

$$V_{n+1}=V_{n}+2$$

$$V_{n+2}=V_{n+1}+2$$

Subtracting the former from the latter, we obtain:

$$V_{n+2}=2V_{n+1}-V_{n}$$

We have the repeated characteristic root:

$$r=1$$

And so the closed form is:

$$V_n=c_1n+c_2$$

Using the initial conditions, we find:

$$V_1=c_1\cdot1+c_2=0\implies c_2=-c_1$$

$$V_2=c_1\cdot2+c_2=2\implies c_1=2,\,c_2=-2$$

Hence:

$$V_n=2(n-1)$$

Thus, we find:

$$V_{513}=2\cdot512=32^2$$

And so:

$$V_{513}=\left(U_{513}-1\right)^2$$

Since we must have $$1\le U_n\forall n\in\mathbb{N}$$, we take the positive root to obtain:

$$U_{513}=32+1=33$$

After I realized I had made an error in the initial conditions while I was away, it turns out Dan was correct after all.

 

[topsquark]

Oooh! I like that method MarkFL. I'll have to remember that trick. (Bow)

-Dan

 

[anemone]

My solution:

Spoiler

If we square the recursion, we may write:

$$\left(U_{n+1}-1\right)^2=\left(U_{n}-1\right)^2+2$$

Let:

$$V_n=\left(U_{n}-1\right)^2$$

And there results:

$$V_{n+1}=V_{n}+2$$

$$V_{n+2}=V_{n+1}+2$$

Subtracting the former from the latter, we obtain:

$$V_{n+2}=2V_{n+1}-V_{n}$$

We have the repeated characteristic root:

$$r=1$$

And so the closed form is:

$$V_n=c_1n+c_2$$

Using the initial conditions, we find:

$$V_1=c_1\cdot1+c_2=0\implies c_2=-c_1$$

$$V_2=c_1\cdot2+c_2=2\implies c_1=2,\,c_2=-2$$

Hence:

$$V_n=2(n-1)$$

Thus, we find:

$$V_{513}=2\cdot512=32^2$$

And so:

$$V_{513}=\left(U_{513}-1\right)^2$$

Since we must have $$1\le U_n\forall n\in\mathbb{N}$$, we take the positive root to obtain:

$$U_{513}=32+1=33$$

After I realized I had made an error in the initial conditions while I was away, it turns out Dan was correct after all.

Hi MarkFL,

Thanks for participating and your answer is correct!(Yes)

My solution:

Spoiler

Note that from $U_1=1$ and working out some

Note that

$(U_{n+1}-1)^2=(U_n-1)^2+2$

So we have

$(U_{n+2}-1)^2=(U_{n+1}-1)^2+2=[(U_n-1)^2+2]+2=(U_n-1)^2+4$

$(U_{n+3}-1)^2=(U_{n+2}-1)^2+2=[(U_n-1)^2+4]+2=(U_n-1)^2+6$ and so on and so forth.

Since we're told that $U_1=1$, the sequence above simplifies to

$(U_2-1)^2=(1-1)^2+2=2=2(1)$,

$(U_3-1)^2=(1-1)^2+4=4=2(2)$,

$(U_4-1)^2=(1-1)^2+6=6=2(3)$$(U_n-1)^2=2(n-1)$

and that gives $(U_{513}-1)^2=2(513-1)=32^2$, i.e. $U_{513}=33$