Fourier expansion for a function:

$$f(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty} e^{- i \omega x}\int_{-\infty}^{+\infty}e^{i\omega t}f(t)dt \, d\omega$$

Newton series expansion of a function:

$$f(x)=\sum_{m=0}^{\infty} \binom {x}m(-1)^{-m} \sum_{k=0}^\infty\binom mk(-1)^{k}f(k)$$

Is there an analogy? Is it possible to make with Newton's series the same things as with Fourier transform? Are the Fourier transform and inverse Fourier transform analoguous to difference delta and summation operators?

  • 2
    $\begingroup$ this looks like the Poisson-Mellin-Newton cycle --- see page 121 of algo.inria.fr/flajolet/Publications/FlSe95.pdf $\endgroup$ Jan 5, 2015 at 8:30
  • $\begingroup$ @Carlo Beenakker well may be, but I fail to see exact math there. $\endgroup$
    – Anixx
    Jan 5, 2015 at 13:23
  • 5
    $\begingroup$ The Newton series is a discrete version of a Taylor series. Fourier series on the unit circle are closely related to Taylor expansion on the unit disk. So one could make a connection by concatenating these two observations, though I don't see what this might be useful for. $\endgroup$
    – Terry Tao
    Jan 5, 2015 at 19:28

2 Answers 2


to give some "exact math", expanding on my comment, here is the Poisson-Mellin-Newton cycle:

from $g(k)$ to $f_n$ via a Newton series: $f_n=\sum_{k=0}^n {n\choose k}(-1)^k g(k)$

from $f_n$ to $f(t)$ via a Poisson generating function: $f(t)=\sum_{n=0}^\infty f_n e^{-t} \frac{t^n}{n!}$

from $f(t)$ to $\hat{f}(s)$ via a Mellin transform: $\hat{f}(s)=\int_0^\infty f(t) t^{s-1}\,dt$

closing the cycle: $\hat{f}(s)=\Gamma(s)g(-s)$.

so if you allow me to substitute your Fourier transform for a Mellin transform, the Poisson generating function allows one to relate that transform to a Newton series, which seems to be the essence of your question.

a related way to make the connection is via the Euler transformation:

We have two involutions, firstly the Newton series (a.k.a. binomial transform)

$$f_n\mapsto g_n=\sum_{k=0}^n {n\choose k}(-1)^k f(k)\mapsto f_n=\sum_{k=0}^n {n\choose k}(-1)^k g(k)$$

and secondly the Euler transformation

$$F(z)\mapsto G(z)=\frac{1}{1-z}F\left(\frac{z}{z-1}\right)\mapsto F(z)=\frac{1}{1-z}G\left(\frac{z}{z-1}\right)$$

The two are related by a Fourier series (or generating function):

$$F(e^{is})=\sum_{n=0}^\infty f_n e^{ins},\;\;G(e^{is})=\sum_{n=0}^\infty g_n e^{ins}$$

Just like for Fourier transforms, there is a correspondence between multiplication and (binomial) convolution, as discussed in this MSE thread.

  • 1
    $\begingroup$ This seems may be related but the exact relation is still unevident to me. I vote +1 because this may be a useful observation. $\endgroup$
    – Anixx
    Jan 5, 2015 at 14:06
  • $\begingroup$ Confusing. The "Poisson gen" step is the Euler trf for e.g.f.s; i.e., $e^{-t}e^{a.t}=e^{-(1-a.)t}= e^{-b.t}$ with $(a.)^n=a_n=f_n$ and $b_n=(1-a.)^n=g_n$. The e.g.f.s are convergent Taylor series on both sides if convergent on one. Not true for the Euler trf for the corresponding o.g.f.s $F(z)$ and $G(z)$; the trf is often invoked to give a definition for the summation of a divergent series in parallel with the Borel method, reflecting $e^{a.t \;D_{y=0}}=e^{-(1-a.)tD_{y=t}}=e^{-b.tD_{y=t}}$ when acting on fcts analytic at both eval points like $e^y$ or $y^n$, but not $y^s$ in general. $\endgroup$ Jan 7, 2015 at 23:01

(This is a revamping of MSE-Q, which has examples of Newton series interpolation for particular functions and the relation to the Mellin transform.)

With $ \; \; \displaystyle \bigtriangledown^{s}_n \; c_n=\sum_{n=0}^{\infty}(-1)^n \binom{s}{n} \; c_n \; \; $ and $\; \; \displaystyle m_j= \int_{0}^{\infty} h(x) \; x^j \; dx \; \; $, formally

$$g(s)=\bigtriangledown^{s-1}_{n} \bigtriangledown^{n}_{j} \frac{m_j}{j!}=\int_0^\infty h(x)\bigtriangledown^{s-1}_{n} \bigtriangledown^{n}_{j} \frac{x^j}{j!} dx=\int_0^\infty h(x) \frac{x^{s-1}}{(s-1)!} \; dx,$$

so since $ \; \; \displaystyle m_j=j!\; g(j+1)=\int_0^\infty h(x) x^j \; dx \; \;$,

$$ g(s+1)=\bigtriangledown^{s}_{n} \bigtriangledown^{n}_{j} g(j+1).$$

Identifying $g(x+1)$ with $f(x)$ in your formula, you can see that your Newton series is the Mellin transform in disguise.

Performing the modified inverse Mellin transform

$h(x)=\displaystyle\frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+ i\infty} \frac{\pi}{sin(\pi s)} g(s) \frac{x^{-s}}{(-s)!} ds$ .

For the Fourier transform, the analogous formula is a sinc function interpolation of a bandlimited function

$$\hat{g}(x) = \int_{-W/2}^{W/2} h(\omega) e^{i 2 \pi x \omega} d\omega.$$


$$\hat{g}(x) = \sum_{n=-\infty}^{\infty} \frac{sin[\pi W\; (n/W-x)]}{ \pi W\;(n/W-x)} \hat{g}(n/W)$$

with $$e^{i2 \pi x \omega}=\sum_{n=-\infty}^{\infty} \frac{sin[\pi W \; (n/W-x)]}{\pi W\; (n/W-x)} e^{i2 \pi n \omega /W} $$

for $-W/2<\omega<W/2$ being the analogue of

$$ \frac{x^{s-1}}{(s-1)!}=\bigtriangledown_{n}^{s-1}\bigtriangledown_{j}^{n} \frac{x^j}{j!}$$

valid for $Real(s)>0$. (This is the basis for the Whittaker-Shannon theorem.)

An aside on a related Newton series and Ramanujan's Master Formula:

Using umbral notation (MSE-Q) for the Taylor series, $f(t)=e^{a.t}$, the normalized Mellin transform formally gives

$$g(s)= \displaystyle \int_{0}^{\infty} e^{-a.t} \frac{t^{s-1}}{(s-1)!}dt$$ $$ = \int_{0}^{\infty} e^{-t}\; e^{(1-a.)t} \frac{t^{s-1}}{(s-1)!}dt =\sum_{m=0}^{\infty} (-1)^m \binom{-s}{m}\sum_{k=0}^m(-1)^k \binom mk a_k \; .$$

Then $f(t)=e^{a.t}$ with $a_n=g(-n)$ since, for $s=-n$, the binomial transform here is an involution and the regularization of the normalized Mellin transform (MT) gives the Dirac delta function or its derivatives in the integrand. Both sides of the equation would converge simultaneously only over restricted values of $s$ as noted in the MO-Q link, but could be analytically continued.

(Use $\binom{s+m-1}{m}=(-1)^m \binom{-s}{m}$ for the MT term-by-term of the Taylor series for $e^{(1-a.)t)}$.)

This is the essence of Ramanujan's Master Formula/Theorem (cf. MO-Q).

So, the Newton interpolation is more akin to the Fourier transform itself; in fact, you can regard the MT here as the interpolation of the coefficients of the Taylor series with the Newton series as one of its avatars.

The inverse MT gives, for appropriate $ \sigma$ when convergent,

$$\displaystyle \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} g(s) \frac{(-t)^{-s}}{(-s)!} ds = \sum_{n=0}^{\infty} g(-n) \frac{t^{n}}{n!} = f(t) \; .$$

  • $\begingroup$ For the last section, identify $g(-x)$ with $f(x)$ to relate it to your formula. $\endgroup$ Jan 8, 2015 at 2:03
  • $\begingroup$ Looking at fractional integroderivatives of Laplace-transform type and their convolutional and related Cauchy-type integrals leads to another interpolation formula. $\endgroup$ Jan 8, 2015 at 4:05
  • $\begingroup$ See also mathoverflow.net/questions/396533/… $\endgroup$ Aug 31 at 19:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.