User:Eric Kvaalen/Notes/An Euler-Maclaurin-like formula

Discussing with Henning Thielemann an asymptotic series which he added for the digamma function, I wonder'd why it had no terms with even powers of x. I suspected that it had something to do with the Euler-Maclaurin formula. After thinking about it a while, I came up with the following formula, similar to the Euler-Maclaurin formula. It works on polynomials, but I haven't seen it elsewhere and I haven't proved to myself that it works for other functions.

The idea is to take the sum of the values at the centres of evenly-spaced intervals as an approximation of the integral and then add corrections. If we apply this to Bernoulli polynomials (whose integrals are zero) the sum is related to the value of the Bernoulli polynomial at 1/2, and we have to correct by that amount. So we have:
 * $$\int_1^nf(x)dx\sim\sum_{k=2}^nf(k-1/2)-\sum_{i=1}\frac{B_{2i}(1/2)}{(2i)!}[f^{(2i-1)}(n)-f^{(2i-1)}(1)]$$

or, putting it the other way around:
 * $$\sum_{k=2}^nf(k-1/2)\sim\int_1^nf(x)dx+\sum_{i=1}\frac{B_{2i}(1/2)}{(2i)!}[f^{(2i-1)}(n)-f^{(2i-1)}(1)]$$

Applying this to the function $$f(x)=1/x$$, we get
 * $$\sum_{k=2}^n\frac 1{k-1/2}\sim\int_1^n\frac 1xdx-\sum_{i=1}\frac{B_{2i}(1/2)}{(2i)!}[(2i-1)!n^{-2i}-(2i-1)!]$$

or
 * $$\sum_{k=2}^n\frac 1{k-1/2}\sim\ln n-\sum_{i=1}\frac{B_{2i}(1/2)}{2i}[n^{-2i}-1].$$

Now, the digamma function is related to the sum on the left. In fact, the sum differs from $$\psi(n+1/2)$$ by some constant. The digamma function is defined such that its difference from the natural logarithm function goes to zero as the argument goes to infinity. So we find that
 * $$\psi(x+1/2)\sim\ln x-\sum_{i=1}\frac{B_{2i}(1/2)}{2i}x^{-2i}.$$

The beginning of this is
 * $$\psi(x+1/2)\sim\ln x+\frac 1{24}x^{-2}$$

which is what one also gets by starting with the asymptotic expansion for $$\psi(x)$$ and transforming it to a series in inverse powers of $x$ for $$\psi(x+1/2).$$

When this expansion, having only even powers, is used to find an expansion for $$\exp(\psi(x+\tfrac 12))$$ one gets only odd powers.

Eric Kvaalen (talk) 13:35, 8 November 2017 (UTC)

Henning has written to me that he thinks my formula can be derived from the original Euler-Maclaurin formula by applying Euler-Maclaurin "to the nodes 0, ½, 1, 1½, 2, ... and then to the integer nodes," and then taking the difference of the two. This seems to be true. We have the Euler-Maclaurin formula:


 * $$\sum_{i=1}^n f(i)\sim\int^n_1 f(x)\,dx+\frac{f(1)+f(n)}2+\sum_{k=1}\frac{B_{2k}}{(2k)!}\left[f^{(2k-1)}(n)-f^{(2k-1)}(1)\right]$$

We can write a second one, but replacing the integral and the derivatives of $$f(x/2)$$ with scaled integral and derivatives of $$f(x):$$


 * $$\sum_{i=2}^{2n}f(i/2)\sim 2\int^n_1 f(x)\,dx+\frac{f(1)+f(n)}2+\sum_{k=1}\frac{B_{2k}}{(2k)!}\frac{f^{(2k-1)}(n)-f^{(2k-1)}(1)}{2^{2k-1}}$$

Subtracting the first from the second gives


 * $$\sum_{i=2}^nf(i-1/2)\sim\int_1^nf(x)dx-\sum_{k=1}\left[1-2^{1-2k}\right]\frac{B_{2k}}{(2k)!}\left[f^{(2k-1)}(n)-f^{(2k-1)}(1)\right]$$

This is the same as my formula above if the values of the Bernoulli polynomials at 1/2 are related to the Bernoulli numbers (the values of the polynomials at 0 or 1) by


 * $$B_{2k}(1/2)=-\left[1-2^{1-2k}\right]B_{2k}.$$

After checking the first couple Bernoulli polynomials, this seems to be the case.

Eric Kvaalen (talk) 15:41, 10 November 2017 (UTC)

In fact, this last relationship is stated at Bernoulli polynomials. Eric Kvaalen (talk) 17:15, 10 November 2017 (UTC)