In the previous entry I gave the positive real-valued solution to the equation,

x

^{n}= x

^{n }

^{– 1 }+ x

^{n }

^{– 2}+ .... + 1, for n = 1, 2, 3, 4, ....9 respectively. Then the ratio in each case of last to preceding term, obtained the corresponding approximation for each of the positive real-valued solutions.

Now the famed golden ratio (r = 1.618033...) relates to the solution of this general equation for n = 2.

For convenience therefore, we can refer to this value as r

_{2.}

In a similar manner, we can refer to the ratios for the 1-step, 2-step, 3-step,..., 9-step sequences as r

_{1}, r

_{2}, r

_{3}, ...,r

_{9 }respectively.

Thus r

_{1}= 1, r

_{2 }= 1.618033...

_{,}r

_{3 }= 1.839286…, r

_{4 }= 1.927561..., r

_{5 }= 1.965948…, r

_{6 }= 1.983585…, r

_{7 }= 1.991966…, r

_{8 }= 1.99603… and r

_{9}= 1.99803…

Then to approximate ζ(s) for s = 1, 2, 3,….,9 we simply substitute the corresponding ratio in the formula 1/(r

_{s }– 1).

In particular when s = 1, ζ(1) → ∞; in like manner 1/(r

_{1 }– 1) = 1/0 likewise → ∞.

When s = 2, ζ(2) = π

^{2}/6 = 1.644934….

Then substituting the value for r

_{2 }in the formula, we get 1/.618033… = 1.618033…

Though this is not of course an exact approximation, it still is fairly good. Indeed concentrating on the significant figures after the decimal point it achieves 95.8% accuracy.

Then when we approximate for higher values of s, the relative accuracy steadily increases.

Indeed ultimately this relative accuracy (for sufficiently high s) → 100%.

So this clearly suggests an intimate connection as between the Riemann zeta function and the extended Fibonacci sequence.

In the following table to illustrate this behaviour I give the values of the Riemann zeta function for s = 1 to 9 respectively with the corresponding approximate values provided through the simple formula using the ratios that apply to the extended Fibonacci sequence.

Finally I give the relative accuracy of the estimates (with respect to the significant figures after the decimal point).

s |
ζ(s) |
Ex. Fibonacci Approximation |
% Accuracy – Significant Figures |

1 | ∞ | ∞ | |

2 | 1.644943 | 1.618033 | 95.8 |

3 | 1.202056 | 1.191488 | 94.8 |

4 | 1.082323 | 1.078096 | 94.9 |

5 | 1.036927 | 1.035252 | 95.5 |

6 | 1.017343 | 1.016688 | 96.2 |

7 | 1.008349 | 1.008099 | 97.0 |

8 | 1.004077 | 1.003985 | 97.8 |

9 | 1.002008 | 1.001973 | 98.3 |

And of course approximations for larger integer values of s can also be provided in the same manner (with the relative accuracy steadily increasing).

However it is not only (positive) integer values of s that can be approximated but all fractional values as well.

To illustrate this, I will explain here how to approximate - for example - the value of ζ(s) for s = 1.5.

Remember the original equation for the 2-step case is given as

,

x

^{2 }– x – 1 = 0.

Again we use the 2-step case. However instead of adding 1 * the last term (in the integer sequence) to 1 * the preceding term, we now add 1 * the last term to .5 * (the previous term).

So for example the first few terms are

0, 1, 1, 1.5, 2, 2.75, 3.75, ….

Now stopping, even at this early stage, the ratio is calculated as 3.75/2.75 = 1.3636.

This approximates to the true solution to the equation given (which approximates to the true solution to the equation x

^{2 }+ bx + c where b = – 1 and c = – .5.

So x

^{2 }– x – .5 = 0 i.e. 2x

^{2 }– 2x – 1 = 0, with the real-valued solution for x = 1.366…

However though it is easy to exactly calculate x in the 2-step case, this would not be true for values of s > 2.

Therefore in these circumstances we would have little option but to resort to the approximate values provided by the corresponding associated sequence of terms.

So using the approximate positive real-valued value for x (as given by the ratio of the last to preceding term in the sequence) we can likewise approximate the value for ζ(1.5) as 1/(1.3636 – 1) = 1/.3636 = 2.75.

Now the correct value for ζ(1.5) = 2.612…

So our approximation - based on just a few terms in a simple series - is 95% accurate!

However as s approaches 1, the Riemann zeta function converges increasingly slowly.

Therefore the simple calculation - based on the extended Fibonacci sequence - can be especially useful in approximating the true answer.

_{}

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