Question

Given, \[{x_1} > a > 0,\]
\[{x_{n + 1}} = \sqrt {{x_n}^2 - 2a{x_n} + 2{a^2},} \]
\[n = 1,2,...\]
Prove that \[\mathop {\lim }\limits_{n \to \infty } {x_n}\]
is convergent and compute its value.

Answer 1

Suppose for a moment it is convergent and has limit L, then
\[ L = \sqrt{L^2 - 2aL + 2a^2} \]
\[ \implies L^2 = L^2 - 2aL + 2a^2 \]
\[ \implies L = a \]

Now, to prove the sequence actually is convergent, show that (xn) is monotonically decreasing and bounded below.

Answer 2

That is great, but can you show me more detail about why xn is decreasing?

Answer 3

That's a challenge for you to show. I recommend trying to show in general for x > a that
\[ \sqrt{x^2 - 2ax + 2a^2} \leq x \]

Answer 4

This statement is true iff
\[ x^2 - 2ax + 2a^2 \leq x^2 \]
which is true iff
\[ a \leq x \]

Answer 5

So to show the sequence is monotonically decreasing, you actually want to show it is bounded below by a.

Answer 6

So now if show that, then you will have that it is indeed bounded below and by the calculation above, it is also monotonically decreasing.

Answer 7

I can see \[{x_2}^2 - {x_1}^2 = 2a(a - {x_1}) < 0\]
\[{x_3}^2 - {x_2}^2 = 2a(a - {x_2}) < 0?\]

Answer 8

and why all that become
\[{x^2} - 2ax + 2{a^2} \leqslant {x^2}\]

Answer 9

If we show for ANY x > a, that
\[ \sqrt{x^2 - 2ax + 2a^2} < x \]
then it will follow that
\[ x_{n+1} < x_n \ \ \hbox{ for all } n\]

Now
\[ \sqrt{x^2 - 2ax + 2a^2} < x \]
is true if and only if, squaring both sides, and as both sides are positive the inequality is preserved,
\[ x^2 - 2ax + 2a^2 < x^2 \]
i.e.,
\[ 2a^2 < 2ax \]
i.e.,
\[ a < x \]

Hence if we can show that \( x_n > a \) for all n, then it follows that the sequence \( (x_n) \) is also monotonically decreasing.

Answer 10

I know x1>a , But why xn > a ?

Answer 11

To show that is true start with an arbitrary number x > a. Show that that implies
\[ \sqrt{x^2 - 2ax + 2a^2} > a --- (*)\]

That being true, it follows by induction that \( x_n > a \) for all n. (And that is because we know that \( x_1 > a \) and the relationship (*) shows that \( x_k > a \implies x_{k+1} > a \) for arbitrary k)

Answer 12

I belive you have try your best to explain the reason. But ..I don't get it..Let me see more book. ..I wish you are my classmate.

Answer 13

ok.

To recap the overall method here:
- show (xn) is bounded below
- show (xn) is decreasing
- then by a standard and very important theorem in analysis, (xn) is a convergent series
- knowing now that (xn) is convergent, write L for it's limit; we can solve for L and find it is a.

Answer 14

Have a look at your lecture notes or text book. I hope/trust there is another example there going through these sorts of steps.

Answer 15

En. I will. Thank you so much.

Answer 16

I got it, JamesJ . haha. Thanks. If I ask other question, please help me, and I promise my question will be more interesting.