Approaching an irrational constant

[Unity+]

As it is known, with a constant such as phi, lim n->infty F_n/F_n-1 = phi, the function can approach but never reach the irrational constant. This can be observed with the following:

 

Phi = lim n->infty F_n/F_n-1 = (1+sqrt(5))/2

Lim n->infty 2F_n = Lim n->infty F_n-1(1+sqrt(5))

 

Turn this into a function displaying the approach to the constant.

 

A(n)=2F_n-F_n-1(1+sqrt(5))

 

Of course, the expected result is to see the function decrease towards 0, but the question I am asking is if this approach towards an irrational constant will always be exponential in nature? If so, is there a way to determine the equation of this decrease without the brute force calculation?

 

Edit: sorry if I didn't use latex. I am on a device that causes problems with it.

Edited August 12, 2014 by Unity+

[mathematic]

http://en.wikipedia.org/wiki/Newton's_method

 

Read the above.

[Unity+]

http://en.wikipedia.org/wiki/Newton's_method

 

Read the above.

 

But how would that apply to this conundrum?

[mathematic]

http://mathworld.wolfram.com/GoldenRatio.html

 

This is more specific.

[Unity+]

http://mathworld.wolfram.com/GoldenRatio.html

 

This is more specific.

That is related to the example, but not an answer to he question.

[mathematic]

If I understand your original question, convergence may or may not be exponential. It depends on the specific sequence.

[Unity+]

If I understand your original question, convergence may or may not be exponential. It depends on the specific sequence.

But is there a way to determine if it will be or not?

[Bignose]

There is a concept of rate of convergence of a series that seems like it would be applicable:

 

http://en.wikipedia.org/wiki/Rate_of_convergence

 

And it is known that different series representation of numbers converge faster or slower.

 

for example

 

[math]\pi = 4 \sum\limits_{k=1}^\infty \frac{(-1)^{k+1}}{2k-1}[/math] converges very slowly (I've seen this called the Gregory series). This one is so poor that is takes more than 300 terms to get pi correct to 2 decimal places.

 

On the other hand the Ramanujan series (actually just one of many he discovered)

 

[math]\frac{1}{\pi} = \frac{2\sqrt{2}}{9801} \sum\limits_{k=0}^\infty \frac{(4k)! (1103+26390k)}{(k!)^4 396^{4k}}[/math] is known to correctly add 8 decimal points of accuracy with every additional term in the series. It is obviously converging faster than the Gregory series.

 

I don't have any proof to say it, but I seriously doubt that all approaches are 'exponential'. The exact rate of convergence would be dependent on the function itself.

 

Edited to fix an error in a formula

Edited August 20, 2014 by Bignose

[Unity+]

I don't have any proof to say it, but I seriously doubt that all approaches are 'exponential'. The exact rate of convergence would be dependent on the function itself.

But I think it would be interesting if there was a function that could be derived from the function that could produce based on ticks how accurate the constant will be at the particular tick.

 

F(t) = accuracy value(i.e, 5 decimals correct).

[Edwina Lee]

The specification of Phi is indeterminate because F(n) was not assigned a value for the 1st value of n.

 

If F(n1) is -ve, then every subsequent term would have alternating sign.

 

If F(n1) is +ve, then every subsequent term would have +ve sign.

 

It is also untrue that Lim n-> ∞ F(n) -> 0. In fact, | Lim n-> ∞ F(n) -> ∞.

 

The original question as I understand asks if the limit of such sequences can be calculated without brut calculation. Well, my answer is it depends on each sequence.

 

In this sequence, we deduced what the limit value is by deduction without brut calculation.

 

Can all sequences be deduced? I don't think so.

[John Cuthber]

The specification of Phi is indeterminate because F(n) was not assigned a value for the 1st value of n.

 

If F(n1) is -ve, then every subsequent term would have alternating sign.

 

If F(n1) is +ve, then every subsequent term would have +ve sign.

 

Are you sure?

I think the ratio of successive terms of the series trends to phi. no matter what the initial terms are.