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In mathematics, the Riesz function is an entire function defined by Marcel Riesz in connection with the Riemann hypothesis, by means of the power series
## Riesz criterion

It can be shown that## Mellin transform of the Riesz function

## Calculation of the Riesz function

# sum_{k=1}^infty frac{(-1)^{k+1}x^k}{(k-1)!} left(sum_{n

1}^infty mu(n)n^{-2k}right)
## Appearance of the Riesz function

## References

- $\{rm\; Riesz\}(x)\; =\; -sum\_\{k=1\}^infty\; frac\{(-x)^k\}\{(k-1)!\; zeta(2k)\}$

- $frac\{x\}\{2\}\; coth\; frac\{x\}\{2\}\; =\; sum\_\{n=0\}^infty\; c\_n\; x^n\; =\; 1\; +\; frac\{1\}\{12\}\; x^2\; -\; frac\{1\}\{720\}x^4\; +\; cdots$

- $F(x)\; =\; sum\_\{n=1\}^infty\; frac\{x^n\}\{c\_\{2n\}(n-1)!\}\; =\; 12x\; -\; 720x^2\; +\; 15120x^3\; -\; cdots$

The values of ζ(2k) approach one for increasing k, and comparing the series for the Riesz function with that for $x\; exp(-x)$ shows that it defines an entire function. The series is one of alternating terms and the function quickly tends to minus infinity for increasingly negative values of x. Positive values of x are more interesting and delicate.

- $operatorname\{Riesz\}(x)\; =\; O(x^e)$

for any exponent e larger than 1/2, where this is big O notation; taking values both positive and negative. Riesz showed that the Riemann hypothesis is equivalent to the claim that the above is true for any e larger than 1/4.

The Riesz function is related to the Riemann zeta function via its Mellin transform. If we take

- $\{mathbf\; M\}(\{rm\; Riesz\}(z))\; =\; int\_0^infty\; \{rm\; Riesz(z)\}\; z^s\; frac\{dz\}\{z\}$

- $int\_0^1\; \{rm\; Riesz\}(z)\; z^s\; frac\{dz\}\{z\}$

- $int\_1^infty\; \{rm\; Riesz\}(z)\; z^s\; frac\{dz\}\{z\}$

From the inverse Mellin transform, we now get an expression for the Riesz function, as

- $\{rm\; Riesz\}(z)\; =\; int\_\{c\; -\; i\; infty\}^\{c+i\; infty\}\; frac\{Gamma(s+1)\}\{zeta(-2s)\}\; z^\{-s\}\; ds$

G.H. Hardy gave the integral representation of $f(x)$ using Borel resummation as

$1-exp(-x)=\; int\_\{0\}^\{infty\}dt\; frac\{f(t)\}\{t\}lfloor\; (frac\{x\}\{t\})^\{1/2\}rfloor$.

The Maclaurin series coefficients of F increase in absolute value until they reach their maximum at the 40th term of -1.753 x 10^{17}. By the 109th term they have dropped below one in absolute value. Taking the first 1000 terms suffices to give a very accurate value for
$F(z)$ for $|z|\; <\; 9$. However, this would require evaluating a polynomial of degree 1000 either using rational arithmetic with the coefficients of large numerator or denominator, or using floating point computations of over 100 digits. An alternative is to use the inverse Mellin transform defined above and numerically integrate. Neither approach is computationally easy.

Another approach is to use acceleration of convergence. We have

- $\{rm\; Riesz\}(x)\; =\; sum\_\{k=1\}^infty\; frac\{(-1)^\{k+1\}x^k\}\{(k-1)!\; zeta(2k)\}$

Using Kummer's method for accelerating convergence gives

- $\{rm\; Riesz\}(x)\; =\; x\; exp(-x)\; -\; sum\_\{k=1\}^infty\; left(zeta(2k)\; -1right)\; left(frac\{(-1)^\{k+1\}\}$

Continuing this process leads to a new series for the Riesz function with much better convergence properties:

- $\{rm\; Riesz\}(x)\; =\; sum\_\{k=1\}^infty\; frac\{(-1)^\{k+1\}x^k\}\{(k-1)!\; zeta(2k)\}$

- $sum\_\{k=1\}^infty\; sum\_\{n=1\}^infty\; frac\{(-1)^\{k+1\}left(x/n^2right)^k\}\{(k-1)!\}=\; x\; sum\_\{n=1\}^infty\; frac\{mu(n)\}\{n^2\}\; expleft(-frac\{x\}\{n^2\}right)$

- $\{rm\; Riesz\}(x)\; =\; x\; left(frac\{6\}\{pi^2\}\; +\; sum\_\{n=1\}^infty\; frac\{mu(n)\}\{n^2\}left(expleft(-frac\{x\}\{n^2\}right)\; -\; 1right)right)$

The above series are absolutely convergent everywhere, and hence may be differentiated term by term, leading to the following expression for the derivative of the Riesz function:

- $\{rm\; Riesz\}\text{'}(x)\; =\; frac\{x\}\; -\; xleft(sum\_\{n=1\}^infty\; frac\{mu(n)\}\{n^4\}\; expleft(-frac\{x\}\{n^2\}right)right)$

- $\{rm\; Riesz\}\text{'}(x)\; =\; frac\{x\}\; +\; xleft(-frac\{90\}\{pi^4\}\; +\; sum\_\{n=1\}^infty\; frac\{mu(n)\}\{n^4\}\; left(1-expleft(-frac\{x\}\{n^2\}right)right)right)$

A plot for the range 0 to 50 is given above. So far as it goes, it does not indicate very rapid growth and perhaps bodes well for the truth of the Riemann hypothesis.

- Titchmarsh, E. C., The Theory of the Riemann Zeta Function, second revised (Heath-Brown) edition, Oxford University Press, 1986, [Section 14.32]

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Last updated on Wednesday June 25, 2008 at 05:43:36 PDT (GMT -0700)

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