1 + 1 + 1 + ⋯ 1 + ⋯ 1 + 1 + ⋯ 1 + ⋯ 1 + 1 + 1 + ⋯ 1 + ⋯ 1 + 1 + ⋯ 1 + ⋯ = ?
Give your answer to 3 decimal places.
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Golden solution!
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Thanks for the compliment!
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Your solution is very correct and i am impressed with you
This solution is incomplete: You need to show convergence.
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How are we going to show convergence here?
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We have to find the limit of the sequence x 0 = 1 and x n + 1 = 1 + x n 1 + x n = 1 + x n , with the iteration function f ( x ) = 1 + x . This function has the sole fixed point c = 2 1 + 5 , as @Lee Care Gene found.
Now f ( x ) is a contraction for x ≥ 0 since f ′ ( x ) = 2 1 + x 1 ≤ 2 1 ; this implies that lim x n = c as claimed. Indeed, by the Mean Value Theorem, we see that ∣ x n + 1 − c ∣ < 2 1 ∣ x n − c ∣ .
You can illustrate this convergence behavior by drawing a cobweb between the graphs of f ( x ) = 1 + x and f ( x ) = x , starting at x 0 = 1 and approaching c from below.
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@Otto Bretscher – i can see that you love cobwebs,and so do i :)
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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
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@Otto Bretscher – I love cobwebs too! They are a wonderful tool for visualising convergence
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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.
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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
@Otto Bretscher is right,without showing convergence,this can't be a valid solution
Nice Solution.
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1 + 1 + 1 + ⋯ 1 + ⋯ 1 + 1 + ⋯ 1 + ⋯ 1 + 1 + 1 + ⋯ 1 + ⋯ 1 + 1 + ⋯ 1 + ⋯ = ?
1 + x 1 + x = x
x + 1 = x
x + 1 = x 2
x 2 = x + 1
x 2 − ( x + 1 ) = x + 1 − ( x + 1 )
x 2 − x − 1 = 0
x = 2 1 ( 1 + 5 ) , 2 1 ( 1 − 5 )
After plugging in, we know x = 2 1 ( 1 − 5 ) .
x = 2 1 ( 1 + 5 )
x ≈ 1 . 6 1 8 (3 D.P.)