Fdtxjr

I was reading this and found on page 6 was a description of a generalization of Taylor series to linear functionals. Reproduced below as

$$ F[phi + lambda] = sum_n=0^infty left[frac1n! intint...int fracdelta ^n F[phi] delta phi(x^0) delta phi(x^1) ... delta phi(x^n-1) dx^0 dx^1 ... dx^n-1
right] $$

I don't understand exactly what the term $$ fracdelta ^n F[phi] delta phi(x^0) delta phi(x^1) ... delta phi(x^n-1) $$

Means.

And let me be extremely concrete. Consider an operator $F: phi rightarrow phi + x^2phi'$, quite literally what this means is that $F: (mathbbR rightarrow mathbbR) rightarrow (mathbbR rightarrow mathbbR) $

And $F:= phi(x) rightarrow phi(x) + x^2 fracddxleft[ phi(x)right] $ Can be viewed as an explicit "form".

So when we compute $$fracdelta Fdelta phi(x)$$

Everything makes sense (since the argument of the $phi$ we are differentiating w.r.t matches the argument of the $phi$ in the original function and the corresponding $fracddx$ is compatible with the $x$ arguments in both cases.) And we use our typical euler lagrange equations.

$$ fracdelta Fdelta phi(x) = fracpartial Fpartial phi(x) - fracddx left[ fracpartial Fpartial left( fracddx[phi(x)] right) ...right] = 1+2x $$

If you make an expression like

$$fracdelta Fdelta phi(x^0)$$

Now it's unclear what you mean, but we can guess. Let us substitute every instance of the string "$x$" with "$x^0$" in $F$ and then this statement can be still well defined. And we conclude that it is equal to

$$ 1 + 2x^0 $$

Now consider:

$$fracdelta^2 Fdelta phi(x^0) delta phi(x^1) $$

This is now getting us into bad territory. Not only do we not have a compatible variable $x$ in either of arguments but we cannot implicitly "guess" what string substitution needs to be made here since any global level string subtitution will render the other derivative to 0. So I am completely lost as to what this could mean.

I'll run through your example and hopefully that makes it clear.

$$phimapsto F[phi]= phi+x^2phi'$$
Vary this
$$phi+deltaphimapsto F[phi]+deltaphi+x^2deltaphi'$$
So we have formally
$$fracF[phi+deltaphi](x)-F[phi](x)deltaphi(y)=fracdeltaphi(x)deltaphi(y)+x^2fracdeltaphi'(x)deltaphi(y)$$

Then we take as our definition that (and more generally would have ignored $mathcal O(deltaphi)$ terms if we have things like $F[phi]=phi^2$)
$$
fracdeltaphi(x)deltaphi(y):=delta(x-y),
$$
the Dirac delta function. Similarly,
$$
fracdeltaphi'(x)deltaphi(y):=delta'(x-y).
$$
One can make this rigorous with compactly supported test functions and the like.

FWIW, in OP's example OP is considering the functional
$$ F[phi]~:=~ phi(x_0) + x_0^2phi^prime(x_0), $$
which depends on the parameter $x^0inmathbbR$.

Then the functional/variational derivative is
$$ fracdelta F[phi]delta phi(x_1)~=~delta(x_0!-!x_1) + x_0^2delta^prime(x_0!-!x_1), $$
which doesn't depend on $phi$. Therefore the higher functional derivatives vanish.