# Can a recursive, continuous integral be approximated with Gauss-Legendre or similar?

Maybe I need to reformulate(?). Suppose there is this simple function:

$$f(x)=\int_a^b{x \text{d}x}$$

If it were to be discretized, there would be losses due to sampling. In order for the errors to be minimized, one can apply Gauss-Legendre, or similar. Now, in this case, it's simple, convert the limits, apply quadrature, since the function is known and can be calculated at any time without the knowledge of past values.

But say you have a function which is recursive, that is, the output, $y$, is a function of the input, $x$ (known all the time), but depends on itself, too, so I am unsure whether $y=f(x)$ or $y=f(x,y)$, because then it can be written as $y=f(x,f(x,y))$, which is $f(x,f(x,f(x,f(...))))$... I'll choose the former, for simplicity:

$$y = f(x)$$ $$y=\frac 1 {b_2}\left[a_2 x+\int^{t_1}_{t_0}{\left(a_1 x-b_1 y+\int^{t_2}_{t_1}{(a_0 x-b_0 y) \text{d}x}\right)\text{d}x}\right]=$$ $$\frac 1 {b_2}\left[a_2 x+\int^{t_1}_{t_0}{\left(a_1 x-b_1 f(x)+\int^{t_2}_{t_1}{(a_0 x-b_0 f(x)) \text{d}x}\right)\text{d}x}\right]\text{?}$$

$t_0$, $t_1$, and $t_2$ are the sampling times. Since there is an integral whithin an integral, the timings differ (as also seen in the code below).

[/edit]

Can Gauss-Legendre still be applied in this case? If not quadrature, maybe other method? If yes, how?

The reason for this is to implement it in C++. This simple test code came up (a is the vector made of $[a_{2,1,0},b_{2,1,0}]$, u is the input signal, y is the output vector, and T is the sampling period):

void f(const std::vector<double> &a, const std::vector<double> &u, std::vector<double> &y, const double &T)
{
int n {static_cast<int>(u.size())};
std::vector<double> y0(n), y1(n);
double a2 {a[0]}, a1 {a[1]}, a0 {a[2]}, b2 {1.0/a[3]}, b1 {a[4]}, b0 {a[5]};
y[0]  = b2*a2*u[0];
y1[1] = T*(a1*u[0] - b1*y[0]);
y[1]  = b2*(a2*u[1] + y1[1]);
for(int i=2; i<n; ++i)
{
y0[i] = T*(a0*u[i-2] - b0*y[i-2]) + y0[i-1];
y1[i] = T*(a1*u[i-1] - b1*y[i-1] + y0[i]) + y1[i-1];
y[i]  = b2*(a2*u[i] + y1[i]);
}
}


Which works, but the period needs to be very small to get a reasonably good approximation of the continuous function, 0.01, or less, which means the vector will have a considerable size, and can be slow to plot.

• See my edits for proper MathJax usage. – Michael Hardy Mar 15 '18 at 15:59
• "If it were to be discretized, there would be losses due to sampling." - recall that an $n$-point Gauss-Legendre quadrature is exact for polynomials of degree $2n-1$ or lower. If your functions are (piecewise) polynomial, you just need to select the right $n$. – J. M. is a poor mathematician Mar 31 '18 at 11:08
• @J.M.isnotamathematician I see, then this could be done, select the appropriate number of points, but I really am not a mathematician, I can't see how to calculate the quadrature when it's a function of itself. I'm sorry if I pollute the air with such trivialities. – a concerned citizen Mar 31 '18 at 11:14
• I actually find your "recursive" notation quite confusing, due in no small part to your omission of the differential: are you, for example, integrating with respect to $x$ or $y$ in the innermost integral? And what are the bounds? (Gauss-Legendre is for definite integration, with a known integration interval, and not indefinite integration, where the result is an antiderivative.) As an aside: your $f(x)$ is a constant, $\frac{b^2-a^2}{2}$. – J. M. is a poor mathematician Mar 31 '18 at 11:17
• In truth, I am in doubt, but I suppose it's $y=f(x)$, that is, a function of the input, which is known at all times. So then, I suppose the integral would be $\int{a*x-b*f(x)\text{d}x}$. And the limits, they would be from T1 to T2, the chosen sampling timings. In the modified question, at the end, I am trying to explain a thought, as childish as it may seem: choose sampling frequency (nr. of points), then apply quadrature between those points. No idea how orthodox it is, but it would have to be a "moving" quadrature, move from sample to sample. Does this mean the order is 2 (=> quad. is 6)? – a concerned citizen Mar 31 '18 at 11:25