Infinite Fractions - 5

Calculus Level 3

1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + = ? \Large \cfrac { 1+\frac { 1+\frac { 1+\cdots }{ \sqrt { 1+\cdots } } }{ \sqrt { 1+\frac { 1+\cdots }{ \sqrt { 1+\cdots } } } } }{ \sqrt { 1+\frac { 1+\frac { 1+\cdots }{ \sqrt { 1+\cdots } } }{ \sqrt { 1+\frac { 1+\cdots }{ \sqrt { 1+\cdots } } } } } } = \, ?

Give your answer to 3 decimal places.


The answer is 1.618.

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1 solution

Lee Care Gene
Feb 15, 2016

1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + = ? \frac { 1+\frac { 1+\frac { 1+\cdots }{ \sqrt { 1+\cdots } } }{ \sqrt { 1+\frac { 1+\cdots }{ \sqrt { 1+\cdots } } } } }{ \sqrt { 1+\frac { 1+\frac { 1+\cdots }{ \sqrt { 1+\cdots } } }{ \sqrt { 1+\frac { 1+\cdots }{ \sqrt { 1+\cdots } } } } } } =?

1 + x 1 + x = x \frac { 1+x }{ \sqrt { 1+x } } =x

x + 1 = x \sqrt { x+1 } =x

x + 1 = x 2 x+1={ x }^{ 2 }

x 2 = x + 1 { x }^{ 2 }=x+1

x 2 ( x + 1 ) = x + 1 ( x + 1 ) { x }^{ 2 }-(x+1)=x+1-(x+1)

x 2 x 1 = 0 { x }^{ 2 }-x-1=0

x = 1 2 ( 1 + 5 ) , 1 2 ( 1 5 ) x=\frac { 1 }{ 2 } (1+\sqrt { 5 } ),\frac { 1 }{ 2 } (1-\sqrt { 5 } )

After plugging in, we know x 1 2 ( 1 5 ) x\neq \frac { 1 }{ 2 } (1-\sqrt { 5 } ) .

x = 1 2 ( 1 + 5 ) x=\frac { 1 }{ 2 } (1+\sqrt { 5 } )

x 1.618 x\approx 1.618 (3 D.P.)

Golden solution!

Saurabh Chaturvedi - 5 years, 4 months ago

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Thanks for the compliment!

Lee Care Gene - 5 years, 4 months ago

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Your solution is very correct and i am impressed with you

Ameya Aglawe - 5 years, 3 months ago

This solution is incomplete: You need to show convergence.

Otto Bretscher - 5 years, 3 months ago

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How are we going to show convergence here?

akash omble - 5 years, 3 months ago

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We have to find the limit of the sequence x 0 = 1 x_0=1 and x n + 1 = 1 + x n 1 + x n = 1 + x n x_{n+1}=\frac{1+x_n}{\sqrt{1+x_n}}=\sqrt{1+x_n} , with the iteration function f ( x ) = 1 + x f(x)=\sqrt{1+x} . This function has the sole fixed point c = 1 + 5 2 c=\frac{1+\sqrt{5}}{2} , as @Lee Care Gene found.

Now f ( x ) f(x) is a contraction for x 0 x\geq 0 since f ( x ) = 1 2 1 + x 1 2 f'(x)=\frac{1}{2\sqrt{1+x}}\leq \frac{1}{2} ; this implies that lim x n = c \lim x_n=c as claimed. Indeed, by the Mean Value Theorem, we see that x n + 1 c < 1 2 x n c |x_{n+1}-c|< \frac{1}{2}|x_n-c| .

You can illustrate this convergence behavior by drawing a cobweb between the graphs of f ( x ) = 1 + x f(x)=\sqrt{1+x} and f ( x ) = x f(x)=x , starting at x 0 = 1 x_0=1 and approaching c c from below.

Otto Bretscher - 5 years, 3 months ago

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@Otto Bretscher i can see that you love cobwebs,and so do i :)

Hamza A - 5 years, 3 months ago

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@Hamza A I only wish I knew how to plot them (even dynamically), as @Agnishom Chattopadhyay has done so beautifully here

Otto Bretscher - 5 years, 3 months ago

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@Otto Bretscher I love cobwebs too! They are a wonderful tool for visualising convergence

Agnishom Chattopadhyay - 5 years, 3 months ago

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@Agnishom Chattopadhyay Sadly, most people on Brillant seem to ignore the issue of convergence when it comes to nested problems. People are content just finding the fixed point(s), which is usually the easy part.

Otto Bretscher - 5 years, 3 months ago

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@Otto Bretscher Actually, not everyone is familiar with the idea that not every sequence converges, for problems such as this, and other sequence problems involving infinte terms. I used to do this mistake myself

Agnishom Chattopadhyay - 5 years, 3 months ago

@Otto Bretscher is right,without showing convergence,this can't be a valid solution

Hamza A - 5 years, 3 months ago

Nice Solution.

A Former Brilliant Member - 5 years, 4 months ago

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