UNSW mathematics competition (1 Viewer)

[Mark576]

This is a practice question from the senior paper:

What is the value of the infinite square root:

root ((((1 + root (((1 + root ((1 + root (1 + ...))))

Any help would be appreciated in solving this problem.

P.S Sorry about the setting out of the problem, but hopefully you get the idea, if not here's a link:

http://www.maths.unsw.edu.au/highschool/teacher/teacherproblems.html (It's the second last problem)

 

[Mark576]

Thanks for that

 

[m.jakaran]

Because it is a repeated root, denote it by x and then place the x inside as the series is infinate. You should then gain a simple quadratic equation.

 

[jet]

 

[Trebla]

For the quadratic solution, we take the positive root because clearly phi is positive with the square roots of positive numbers.

 

[3unitz]

the problem with this proof is that it assumes root ((((1 + root (((1 + root ((1 + root (1 + ...)))) converges to a finite number phi. for example, how do we know it doesnt shoot off to infinity instead?

allow me to illustrate my point by considering a similar "repetition":

((( (...) - 1)^2 - 1)^2 - 1)^2

we could define this as the limit of the sequence a_n with the following recursion formulae:

a₀ = 0

a_n+1 = (a_n - 1)^2

if we assume this sequence has a finite limit L then by limit laws:

L = (L - 1)^2

which implies:

L = [3 +/- root(5)] / 2

L ~ 2.618, L ~ 0.382

this is nonsense. if instead we look at each iteration, we observe that the sequence oscillates between 0 and 1 and does not converge:

a₀ = 0

a₁ = 1

a₂ = 0
.
.
.

this example shows that its possible to obtain an incorrect limit just by simply assuming the sequence converges. returning back to the infinite square root problem, we must first be sure that it does indeed have a finite limit before we try to manipulate any limit algebra.

to do this we have to prove 2 points:

(a) that the sequence is monotonically increasing (a_n+1 > a_n for all n) and thus cannot oscillate.

(b) that there is a constant number bigger than any value of the sequence for all n (like an asymptote which "blocks" the sequence from diverging to infinity)

if we can show that (a) and (b) are true, we have effectively shown the sequence must converge to a finite limit.

define the infinite square root as the limit of the sequence a_n with the following recursion formulae:

a₀ = 0

a_n+1 = root(1 + a_n) --------(1)

by inspection: a_n < a_n+1 ((a) is true)

rearranging (1) for a_n and subbing into this inequality yields:

a_n+1^2 - 1 < a_n+1

a_n+1^2 - a_n+1 - 1 < 0

this implies a_n+1 < [1 + root(5)]/2 (using the quadratic equation)

a_n < a_n+1 < [1 + root(5)]/2

(b) is true as the sequence is always less than the constant [1 + root(5)]/2

therefore, the sequence must converge to some finite number, phi, and we can only now apply limit laws:

phi = root(1 + phi)

phi = [1 + root(5)]/2

 

[HalcyonSky]

the problem with this proof is that it assumes root ((((1 + root (((1 + root ((1 + root (1 + ...)))) converges to a finite number phi. for example, how do we know it doesnt shoot off to infinity instead?

allow me to illustrate my point by considering a similar "repetition":

((( (...) - 1)^2 - 1)^2 - 1)^2

we could define this as the limit of the sequence a_n with the following recursion formulae:

a₀ = 0

a_n+1 = (a_n - 1)^2

if we assume this sequence has a finite limit L then by limit laws:

L = (L - 1)^2

which implies:

L = [3 +/- root(5)] / 2

L ~ 2.618, L ~ 0.382

this is nonsense. if instead we look at each iteration, we observe that the sequence oscillates between 0 and 1 and does not converge:

a₀ = 0

a₁ = 1

a₂ = 0
.
.
.

this example shows that its possible to obtain an incorrect limit just by simply assuming the sequence converges. returning back to the infinite square root problem, we must first be sure that it does indeed have a finite limit before we try to manipulate any limit algebra.

to do this we have to prove 2 points:

(a) that the sequence is monotonically increasing (a_n+1 > a_n for all n) and thus cannot oscillate.

(b) that there is a constant number bigger than any value of the sequence for all n (like an asymptote which "blocks" the sequence from diverging to infinity)

if we can show that (a) and (b) are true, we have effectively shown the sequence must converge to a finite limit.

define the infinite square root as the limit of the sequence a_n with the following recursion formulae:

a₀ = 0

a_n+1 = root(1 + a_n) --------(1)

by inspection: a_n < a_n+1 ((a) is true)

rearranging (1) for a_n and subbing into this inequality yields:

a_n+1^2 - 1 < a_n+1

a_n+1^2 - a_n+1 - 1 < 0

this implies a_n+1 < [1 + root(5)]/2 (using the quadratic equation)

a_n < a_n+1 < [1 + root(5)]/2

(b) is true as the sequence is always less than the constant [1 + root(5)]/2

therefore, the sequence must converge to some finite number, phi, and we can only now apply limit laws:

phi = root(1 + phi)

phi = [1 + root(5)]/2

 

[alakazimmy]

LOL!! 3unitz just piled on a shitload of university maths.

Do MATH1131/1141/1151 at UNSW if you want to learn all this convergence, divergence crap for sequences.

 

[m.jakaran]

the problem with this proof is that it assumes root ((((1 + root (((1 + root ((1 + root (1 + ...)))) converges to a finite number phi. for example, how do we know it doesnt shoot off to infinity instead?

allow me to illustrate my point by considering a similar "repetition":

((( (...) - 1)^2 - 1)^2 - 1)^2

we could define this as the limit of the sequence a_n with the following recursion formulae:

a₀ = 0

a_n+1 = (a_n - 1)^2

if we assume this sequence has a finite limit L then by limit laws:

L = (L - 1)^2

which implies:

L = [3 +/- root(5)] / 2

L ~ 2.618, L ~ 0.382

this is nonsense. if instead we look at each iteration, we observe that the sequence oscillates between 0 and 1 and does not converge:

a₀ = 0

a₁ = 1

a₂ = 0
.
.
.

this example shows that its possible to obtain an incorrect limit just by simply assuming the sequence converges. returning back to the infinite square root problem, we must first be sure that it does indeed have a finite limit before we try to manipulate any limit algebra.

to do this we have to prove 2 points:

(a) that the sequence is monotonically increasing (a_n+1 > a_n for all n) and thus cannot oscillate.

(b) that there is a constant number bigger than any value of the sequence for all n (like an asymptote which "blocks" the sequence from diverging to infinity)

if we can show that (a) and (b) are true, we have effectively shown the sequence must converge to a finite limit.

define the infinite square root as the limit of the sequence a_n with the following recursion formulae:

a₀ = 0

a_n+1 = root(1 + a_n) --------(1)

by inspection: a_n < a_n+1 ((a) is true)

rearranging (1) for a_n and subbing into this inequality yields:

a_n+1^2 - 1 < a_n+1

a_n+1^2 - a_n+1 - 1 < 0

this implies a_n+1 < [1 + root(5)]/2 (using the quadratic equation)

a_n < a_n+1 < [1 + root(5)]/2

(b) is true as the sequence is always less than the constant [1 + root(5)]/2

therefore, the sequence must converge to some finite number, phi, and we can only now apply limit laws:

phi = root(1 + phi)

phi = [1 + root(5)]/2

wow, do you do maths at uni?