# Infinite surds - expression for which the exact value is an integer

• Aug 25th 2008, 01:35 AM
starvin-marvin
Infinite surds - general statement for integer exact surd values
For the surd

$\displaystyle a_n = \sqrt{k + \sqrt{k + \sqrt{k + \sqrt{k + ....... \infty }}}}$

therefore

$\displaystyle x = ({1+\sqrt{1+4k}}){/2}$

The task is: What's a general statement which represents values for k which make the exact value an integer.

What does the question mean by 'general statement'? what form is my answer supposed to be in?
• Aug 25th 2008, 02:24 AM
mr fantastic
Quote:

Originally Posted by starvin-marvin
For the infinite surd
$\displaystyle a_3 = \sqrt{k + \sqrt{k + \sqrt{k + \sqrt{k}}}}$

$\displaystyle a_n = \sqrt{k + \sqrt{k + \sqrt{k + \sqrt{k + ....... \infty }}}}$

$\displaystyle a_n+1 = \sqrt{k + a_n}$

therefore

$\displaystyle a_\infty = \sqrt{k + a_\infty}$

let $\displaystyle a_\infty = x$

$\displaystyle x = \sqrt{k + x}$

$\displaystyle x^2 -x - k = 0$

$\displaystyle x = ({1+\sqrt{1+4k}}){/2}$

What is the general statement that represents all values of k for which the expression is an integer.

This is for your IB project, right?

• Aug 25th 2008, 02:28 AM
Moo
Quote:

Originally Posted by starvin-marvin
For the infinite surd
$\displaystyle a_3 = \sqrt{k + \sqrt{k + \sqrt{k + \sqrt{k}}}}$

$\displaystyle a_n = \sqrt{k + \sqrt{k + \sqrt{k + \sqrt{k + ....... \infty }}}}$

$\displaystyle a_n+1 = \sqrt{k + a_n}$

therefore

$\displaystyle a_\infty = \sqrt{k + a_\infty}$

let
$\displaystyle a_\infty = x$

$\displaystyle x = \sqrt{k + x}$

$\displaystyle x^2 -x - k = 0$

$\displaystyle x = ({1+\sqrt{1+4k}}){/2}$

What is the general statement that represents all values of k for which the expression is an integer.

1+4k has to be a perfect square. Why ? Because there's a theorem saying that $\displaystyle \sqrt{n}$ is an irrational iff n is not a perfect square.
Now, if 1+4k is a perfect square, it will be an odd integer. And the square root of an odd integer is an odd integer. Added to 1, it'll give an even integer, which, divided by 2, will be an integer.
• Aug 25th 2008, 03:27 AM
starvin-marvin
Thanks for the reply. So is that a 'general statement'? Is it possible to create some kind of formula like the http://www.mathhelpforum.com/math-he...dbd98fec-1.gif used, which only works for values of k that make x an integer?

So is all you need to completely satisfy that goal simply state the conditions for k, that 4k + 1 has to be a perfect square?
• Aug 25th 2008, 07:22 AM
Soroban
Hello, starvin-marvin!

Quote:

Is it possible to create some kind of formula like the http://www.mathhelpforum.com/math-he...dbd98fec-1.gif used,
which only works for values of k that make x an integer?

The exact question is: Find the general statement that represnts all the values of $\displaystyle k$
for which the expression is an integer.

So is all you need to completely satisfy that goal
simply state the conditions for $\displaystyle k$, that $\displaystyle 4k + 1$ is a perfect square? . . . . yes

We have: .$\displaystyle 4k + 1 \:=\:b^2$, for some integer $\displaystyle b$

. . Then: .$\displaystyle k \:=\:\frac{b^2-1}{4}\;\;{\color{blue}[1]}$

We see that: $\displaystyle b^2-1 \text{ is even }\quad\Rightarrow\quad b^2\text{ is odd }\quad\Rightarrow\quad b\text{ is odd}$

Let: $\displaystyle b \:=\:2c+1$

Then [1] becomes: .$\displaystyle k \:=\:\frac{(2c+1)^2-1}{4} \:=\:c^2 + c \quad\Rightarrow\quad k \:=\:c(c+1)$

Hence, $\displaystyle k$ must be the product of two consecutive integers.

Then: .$\displaystyle x \;=\;\frac{1 + \sqrt{1 + 4k}}{2} \;=\;\frac{1 + \sqrt{1+4c(c+1)}}{2} \;=\;\frac{1 + (2c+1)}{2} \;=\;c+1$, an integer.

• Aug 25th 2008, 02:34 PM
mr fantastic
Quote:

Originally Posted by Soroban
Hello, starvin-marvin!

We have: .$\displaystyle 4k + 1 \:=\:b^2$, for some integer $\displaystyle b$

. . Then: .$\displaystyle k \:=\:\frac{b^2-1}{4}\;\;{\color{blue}[1]}$

We see that: $\displaystyle b^2-1 \text{ is even }\quad\Rightarrow\quad b^2\text{ is odd }\quad\Rightarrow\quad b\text{ is odd}$

Let: $\displaystyle b \:=\:2c+1$

Then [1] becomes: .$\displaystyle k \:=\:\frac{(2c+1)^2-1}{4} \:=\:c^2 + c \quad\Rightarrow\quad k \:=\:c(c+1)$

Hence, $\displaystyle k$ must be the product of two consecutive integers.

Then: .$\displaystyle x \;=\;\frac{1 + \sqrt{1 + 4k}}{2} \;=\;\frac{1 + \sqrt{1+4c(c+1)}}{2} \;=\;\frac{1 + (2c+1)}{2} \;=\;c+1$, an integer.

@starvin-marvin: Since this was for your IB portfolio, I hope you reference all the material in this thread accordingly rather than passing it off as your own work. Particularly the above post, which gives the complete solution.

I hope there are other parts to this investigation which will allow at least some of the portfolio to reflect your own work.
• Aug 25th 2008, 10:16 PM
starvin-marvin
Yes, I came to the conclusion through my own working that k must be the product of two consecutive numbers. I was just wondering whether when it asks for a 'general statement' wether it wants a written description for the conditions of k that make x an integer, or some formula like this one: http://www.mathhelpforum.com/math-he...dbd98fec-1.gif which only produces integer values for x from certain values of k. I was thinking of substituting in one of these conditions for k, which is
$\displaystyle k = ab$ (where a and b are consecutive)
$\displaystyle b = a + 1$
therefore: $\displaystyle k = a^2 + a$
• Nov 20th 2008, 09:24 PM
CaptainBlack
You go from (note the typo in the original):

$\displaystyle a_{n+1}=\sqrt{k+a_n}$

to the next line by observing that if this process converges $\displaystyle a_{n+1}$ and $\displaystyle a_n$ become indistinguishable for large $\displaystyle n$, so:

$\displaystyle \lim_{n \to \infty}a_{n+1}=\lim_{n \to \infty }\sqrt{k+a_n}=\sqrt{k+\lim_{n \to \infty}(a_n)}$

Or using $\displaystyle a_{\infty}$ to represent $\displaystyle \lim_{n \to \infty}a_n=\lim_{n \to \infty}a_{n+1}$ we have:

$\displaystyle a_{\infty}=\sqrt{k+a_{\infty}}$

CB
• Mar 28th 2009, 10:59 AM
mathchic1234
I am still a little confused how u got http://www.mathhelpforum.com/math-he...dbd98fec-1.gif from the previous step. could you possible explain it to me?
• Mar 28th 2009, 12:39 PM
mr fantastic
Quote:

Originally Posted by mathchic1234
I am still a little confused how u got http://www.mathhelpforum.com/math-he...dbd98fec-1.gif from the previous step. could you possible explain it to me?

This thread is closed since as far as I'm aware the IB portfolio is meant to be the work of the student, not the work of others. Ditto here: http://www.mathhelpforum.com/math-he...tml#post289914