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The convolution isWe claim that the $n$-fold convolution isThis is true for $n=1$. If it holds for $n-1$ and $x\ge 0$, thenand the convolution vanishes for $x<0$. This is the gamma density from repeated exponential convolution.

The Fourier transform is

The convolution theorem statesIndeed, Fubini and $u=x-y$ giveSince $\hat{F}(k)=(1+ik{)}^{-1}$, induction immediately verifies

For Parseval identity, take $g(x)=\overline{f(-x)}$. ThenUsing Fourier inversion at zero, which follows from the supplied delta identity,and the convolution theorem gives

Apply this to $F_{n+1}$. Sinceone obtains the Even rational Parseval integral

For a variation $y+\epsilon \eta$,Integration by parts givesFor fixed endpoints, $\eta (0)=\eta (x_0)=0$. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equationA solution makes the first variation vanish for every admissible variation, so it is a stationary candidate; whether it is a minimum or maximum is decided by higher variations. If endpoint values are free, $\eta$ is arbitrary there, and the boundary term instead vanishes under the natural conditionsThus the same Euler-Lagrange solution is stationary for all free-endpoint variations. These are the natural boundary conditions for a free endpoint.

Forthe two equations areSet $u=y+z$ and $v=y-z$. ThenThe conditions $y(0)=z(0)=0$ giveand hence the most general solution is

Free conditions at $x_0$ are $y^{\prime }(x_0)=z^{\prime }(x_0)=0$. Adding and subtracting them givesThus $A=0$. The zero solution exists for every $x_0$, while nonzero solutions exist precisely whenFor those values they form the one-parameter familywhere $C$ is arbitrary. This is the free-endpoint normal mode of a coupled variational functional.

For $0<a<2$, the beta-integral evaluation givesDifferentiation under the integral sign near $a=1$ is justified by domination at zero and infinity. ThereforeWriting $a=1+t$ givesSincethe logarithmic moments of the Cauchy kernel yieldThe same expansion gives $I^{\prime \prime }(1)={\pi }^3/8$, agreeing with the supplied identity.

For distinct nodes, define the divided difference byThis is the coefficient of $x^k$ in the Lagrange interpolation polynomial through the first $k+1$ data points.

Let $p_k$ denote that interpolating polynomial. The difference $p_k-p_{k-1}$ vanishes at $x_0,\dots ,x_{k-1}$, soThe Lagrange formula shows that the leading coefficient of $p_k$ is $f[x_0,\dots ,x_k]$, whereas $p_{k-1}$ has degree at most $k-1$. Hence $c_k=f[x_0,\dots ,x_k]$. Starting with $p_0=f(x_0)$ and iterating gives the Newton interpolation polynomial

The divided-difference recurrence isFor three nodes, the triangular table iswhere each entry in a new column uses the two adjacent entries to its left. There are $n$ first differences, $n-1$ second differences, and so on, each requiring one division. The exact total is