Golden ratio

Calculus Level 2

Find the value of

1 + 1 1 + 1 1 + 1 1 + 1 \large 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{\ddots }}}}


The answer is 1.61803.

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1 solution

Nikhil Raj
May 30, 2017

Let, x = 1 + 1 1 + 1 1 + 1 1 + 1 x = 1 + 1 x x = x + 1 x x 2 = x + 1 x 2 x 1 = 0 x = 1 + 5 2 = 1.61803 x = 1 + \dfrac{1}{1 + \dfrac{1}{1 + \dfrac{1}{1 + \dfrac{1}{\ldots}}}} \\ \Rightarrow x = 1 + \dfrac{1}{x} \\ \Rightarrow x = \dfrac{x +1}{x} \\ \Rightarrow x^2 = x + 1 \\ \Rightarrow x^2 - x -1 = 0 \ \\ \therefore x = \dfrac{1 + \sqrt{5}}{2} = \color{#E81990}{\boxed{1.61803}}

FYI You do not need to restrict to "positive" value. Do you know why?

Calvin Lin Staff - 4 years ago

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Can you explain me clearly.

Nikhil Raj - 4 years ago

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If we take negative sign then the answer would be negative but 1+1÷(1+1÷(1× ....)...) Is positive

genis dude - 4 years ago

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@Genis Dude I got it! cool!

Nikhil Raj - 4 years ago

@Genis Dude Not just that.

Note that the implication sign is only in 1 direction. What this means is that we only have a necessary condition - Any limit must satisfy the given equation. This condition is not sufficient, meaning that any solution to the given equation may not be the limit.

Thus, you're still missing the step of "Justifying that we indeed have found the limit", as opposed to "Well, we only have 1 possible candidate left ...". So, there are two ways to proceed.
1. Showing that 1 + 5 2 \frac{ 1 + \sqrt{5} } { 2} is indeed the limit.
2. Prove that the limit exists, and since we only have 1 candidate, that candidate must be the limit.

Calvin Lin Staff - 4 years ago

I just don't understand how this is calculus.

Zach Abueg - 4 years ago

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See my comments. The partial fraction is defined as the limit of truncated terms.

Yes, you can use algebra to solve for it, as Nikhil done above. However, you've not determined the value rigorously. There are instances where the algebraic solution has to be rejected, and we conclude that "the limit does not exist".

Calvin Lin Staff - 4 years ago

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Oh, interesting! I didn't think about it that way. Thanks for showing me!

Zach Abueg - 4 years ago

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@Zach Abueg As an example of such a scenario, try this sequence. 1 , 2 , 3 .

Calvin Lin Staff - 4 years ago

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@Calvin Lin Makes sense; those definitely enlighten how I see the concept now. Thanks!

Zach Abueg - 4 years ago

1 + 5 2 1.61083 \displaystyle \frac { 1 + \sqrt { 5 } } { 2 } \ne 1.61083 .

. . - 3 weeks, 5 days ago

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