– 1 and – 1 and – 2 respectively.

However it is not very fruitful to attempt to continue in estimating ζ(s) for further negative values of s in this somewhat piecemeal manner.

Some years ago, I formulated a number of approximation formulae for Negative Odd Integers of the Riemann Zeta Function for making such calculations. These are basically of the form that based on one known value an other can then be approximated.

One such formula is given as:

**ζ (s**

**– 2)/ζ(s)**

**→**

**.1 + 11k/2**

**π**

^{2 }**+ k(k**

**– 3)/**

**π**

^{2}
Now for this calculation, we use absolute values of

**ζ(s - 2)**and**ζ(s)**and**k**in the formula is given as**k =**–**(s + 1)/2.**

So when

**s =****– 1**,**k = 0.**
Therefore

**ζ****– 3)/ζ(****– 1**)**→****.1**
So

**,**given that**ζ(****– 1**) =**1/12 (**in absolute term**s),**therefore**ζ****– 3)****→****1/120.**
However the perfect accuracy in this case represents beginner's luck, as approximations generally will not be exact.

For example if we were to continue to attempt to calculate

**ζ****– 5)**from**ζ****– 3)**,**k =**

**– (**

**– 3 + 1)/2 = 1.**
Therefore

**ζ****– 5)****/ζ(****– 3**)**→****.****1 + 11/2****π**^{2}**–2****/****π**^{2 }**= .45462.**
So the absolute value of

**ζ****– 5)****→****.45462(1/120)****→ 1/264.**
Though not exact, this approximates well the true value

In fact, a more accurate approximation can come from using our original approximation for

This would then give the approximation for

And we can also use this formula for non-integer values.

For example let us attempt to calculate the value of

Therefore

So in absolute terms**1/252**and the relative approximation steadily improves for larger values of s.In fact, a more accurate approximation can come from using our original approximation for

**ζ(****i.e.****– 1**)**–****.086427**.This would then give the approximation for

**ζ****– 3)**as**–****.0086427**and the corresponding approximation for**ζ****– 5)**as**.003929... = 1/254.5**in absolute terms which is 99% accurate.And we can also use this formula for non-integer values.

For example let us attempt to calculate the value of

**ζ****– 2.5)**from**our already obtained approximation for****ζ****– 0.5)**i.e.**–****.207879.****k****–****(s + 1)/2 =****– .25**Therefore

**ζ****– 2.5)****/ζ(****– 0.5**)**→****.****1****–2.75**/2**π**^{2}**+ .812****5/****π**^{2 }**= .04300**

**ζ****– 2.5)**

**→ .043*******

**207879 = .008173.**

This compares well with the correct answer

**.008516...**(96% relative accuracy).

So once we have the means of approximating all values for

**in the range for s from**

**ζs****)****0**to

**then we can in principle ultimately approximate**

**– 2,****for all negative values of s using this formula.**

**ζs****)**

**However, there are other methods of approximation, that I suggested at that time.**

Once again where s is a negative odd number, this alternative formula can be used,

**{│ζ(s - 2k)}**

^{1/(1 – s + 2k)}│}/{│ζ(s)}^{1/(1 – s)}│}^{ }**→**

**(1 - s + 2k)/(1 – s)**

When we confine ourselves to negative integer values of s, the value for

**k = 1**.
Therefore in this special case,

**{│ζ(s - 2)}**

^{1/(3 – s)}│}/{│ζ(s)}^{1/(1 – s)}│}^{ }**→**

**(3 - s)/(1 – s)**

This is especially useful for larger absolute values of s.

For example,

**given****knowledge of****│**ζ**– 13)****│**=**let us consider the approximation that can be obtained for****1/12,****.****ζ****– 15)**
So

**{│ζ(- 15)}**^{1/16}│}**/{│ζ(****- 13****)}**^{1/14}│}**→ 16****/14**
Thus

**{│ζ(- 15)}**^{1/16}│}**→ 16****/14 *****{│ζ(****- 13****)}**^{1/14}│}**→ 16/14 * .837366 = .956990**

Therefore

This

This still is not very exact. However the relative accuracy of the approximation continually improves for larger absolute values of s

**│****ζ(-****15****)****│****→.4949**This

**compares with the true value =****3617/8160 = .4434...**This still is not very exact. However the relative accuracy of the approximation continually improves for larger absolute values of s

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