nested radicals

Call the Feds!

We’ve Got Nested Radicals!

Alan Craig

F. Lane Hardy Seminar

11-3-08

What are Nested Radicals?

2222

222

22

Examples:

We could keep this up forever!

222222

If we did, what would we get?

222222 = ?

Let’s work up to it.What are the values of these expressions?

? 22222

? 2222

? 222

? 22

? 2

Let’s work up to it.What are the values of these expressions?

? 22222

? 2222

? 222

? 22

1.414 2

What are the Values?

? 22222

? 2222

? 222

1.848 22

1.414 2


? 22222

? 2222

1.962 222

1.848 22

1.414 2


? 22222

1.990 2222

1.962 222

1.848 22

1.414 2


1.998 22222

1.990 2222

1.962 222

1.848 22

1.414 2

What value is this sequence of numbers approaching?

1.998

1.990

1.962

1.848

1.414

Now what do you think the value of this infinite nested radical is?

? 222222

1.998 22222

1.990 2222

1.962 222

1.848 22

1.414 2

You’re Right!

2 222222

Let’s see an example of where an infinite nested radical could arise.

Warning: Brief Excursion into Trigonometry!

Trigonometry

Half-Angle Formula

• We will use the half-angle formula for cosine to take another look at this sequence and its limit.

2

cos1

2cos

Let’s use the formula to find .

so ,2

2

4cos Now,

24

cos1

8cos

8cos


222

1

24

cos1

8cos

8cos

Let’s rationalize the last expression by multiplying numerator and denominator by 2.


4

22

22

222

1

222

1

24

cos1

8cos

8cos


2

22

4

22

222

1

24

cos1

8cos

8cos


2

22

4

22

222

1

24

cos1

8cos

8cos

Now multiply both sides by 2.


228

cos2

2

22

4

22

222

1

24

cos1

8cos

8cos

Repeatedly using the ½ angle formula:

32cos22222

16cos2222

8cos222


32cos22222

16cos2222

8cos222

As the angle gets smaller and smaller approaching 0, what value is the cos() approaching?


32cos22222

16cos2222

8cos222

Recall cos(0) = 1, so

2 cos() is approaching 2 as approaches 0.


2120cos22222

0 as 1cos

That is,

That’s all the trigonometry for this session.

We have shown in two different ways

that the equation ‘ought’ to be true

2 222222

To Recap:

Now let’s ‘prove’ it.

2 222222

Set x equal to the expression.

222222x

Square both sides.

2222222x

Subtract the original equation from the squared equation.

222222

2222222

x

x

2

222222

222222

2

2

xx

x

x


Now solve the equation.

22 xx

Solve the equation.

02

22

2

xx

xx

Solve the equation.

0)1)(2(

02

22

2

xx

xx

xx

Solve the equation.

2

0)1)(2(

02

22

2

x

xx

xx

xx

Why did we not use x = -1?

So

2 222222

What about?

333333

Does

333333

= 3 ???

Using the same process as before, we get

033

333333

333333

22

2

xxxx

x

x

Recall the Quadratic Formula

a

acbbxcbxax

2

40

22

032 xx• We have

• So a = 1, b = -1, and c = -3 and

2

131

)1(2

)3)(1(4)1()1( 2

x

x

So, No, we do not get 3

3.2333333

so ,3.22

131032

xxxx

Let’s ask a slightly different question.

• Is there a positive integer a, such that if we replace 3 under the nested radical with a, the nested radical will equal 3?


• That is, is there an a that makes the equation below true?

? 3 aaaaaa


• That is, is there an a that makes the equation below true?

• Yes! And we are going to find it.

? 3 aaaaaa


axx

aaaaaax

aaaaaax

2

2

Finding a

2

41102 a

xaxx

(Using the quadratic formula)

Finding a

We want x = 3, so

2

411 ax

32

411

a

Finding a

641132

411a

a

Finding a

2541541

641132

411

aa

aa

Finding a

6

2541541

641132

411

a

aa

aa

So we have shown that

3666666

Now let’s generalize our result.

• ‘Prove’ that for any integer k > 1, there is a unique positive integer a, such that

kaaaaaa

Note: The following is not a true mathematical proof of this theorem (which would use limits of bounded, monotonically increasing sequences) but does suggest the core reasoning and result of such a proof.

Finding a

12412

411

kak

a

Finding a

14441

12412

411

2

kka

kaka

Finding a

14441

12412

411

22

kkakka

kaka

Finding a

)1(

14441

12412

411

22

kka

kkakka

kaka

We have shown that

For any integer k > 1, there is exactly one integer a = k (k - 1), such that

kaaaaaa

We have shown that

For any integer k > 1, there is exactly one integer a = k (k - 1), such that

kaaaaaa

That is, every integer can be represented as an infinite nested radical!

Example: k = 4

4121212121212

1234)1( kka

5202020202020

2045)1( kka

Example: k = 5

Alternatively, we might have noticed that we need to solve

in such a way that we get two numbers that multiply to make a and subtract to make 1. Further, one of the numbers must be k. (Why?) Thus, the other number must be k - 1 and a must be k (k - 1).

Another Way

02 axx

That is

)1(

1

,1 and

:other the and numbers theof one be Let

2

2

kkkka

kakk

ak

k

ah

hkahk

hk

The END?

The END?

No!

This is way too much fun!

Let’s Kick it Up a Notch!

abababababa

abababababa

Note that what we did before was a special case of this expression with b = 1.


kabababababa

For each integer k > 1, there are exactly k - 1 pairs of integers a and b, 0 < b < k, that satisfy this equation. Further, ).( bkka


As before, square the equation.

abababababab

ax

abababababax

2

2

But before we subtract the original equation from the squared equation, we must isolate the radical (so that it will subtract away).

Now subtract.

02

2

xb

ax

abababababax

abababababab

ax

Now subtract.

00 22

2

abxxxb

ax

abababababax

abababababab

ax

We will solve this by factoring now but keep it in mind for later.

For integer solutions of

we need two integers that multiply to make a and have a difference of b. One of the numbers must be k, so the other is k - b. Thus,

Factor

02 abxx

)( bkka

There are exactly k – 1 such pairs a and b:

(k – 1) Pairs

1 1

2 2

3 )3(

2 )2(

1 )1(

)(

kk

kk

kk

kk

kk

bbkka

(difference)

Recall t

hat 0 < b < k

If k = 4, the k – 1 = 3 pairs a and b are:

Example: k = 4

3 414

2 824

1 1234

ba

Example: k = 4

482828282828

4121212121212

443434343434

One Last Thought

ba

b

ab

ab

ab

Consider this continued fraction:

ba

b

ab

ab

abx

Suppose it converges to x, then

ba

b

ab

ab

abx

Notice the shaded area is also x

Rewriting the continued fraction

x

abx

ba

b

ab

ab

abx

See what we get!

02

abxx

x

abx

ba

b

ab

ab

abx

Does this look familiar?

Yes, these are equal!!!

ababababa

ba

b

ab

ab

ab

In particular, set a = b = 1.

???111111111

11

1

11

11

11

The Golden Ratio

2

51111111111

11

1

11

11

11

(But that’s another F. Lane Hardy talk.)

EndEndEndEndEnd

?Reference

Zimmerman, S., & Ho, C. (2008). On infinitely nested radicals. Mathematics Magazine, 81(1), 3-15.

nested radicals

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