PARTIAL NORMAL MOVEOUT

The case of the N^th partial normal moveout that is not a truncated Taylor series is easier, so let us examine it first. Define now the N^th partial normal moveout by  

$$\tau_N(t) = \sqrt{t^2-{x^2 \over Nv^2}}.$$ (5)

We can see that the truncated Taylor series of equation (5) is equation (2). Now we will prove that

$$T_N(t)=\tau_N \tau_N \cdots \tau_N(t)=\sqrt{t^2-{x^2 \over v^2}}$$ (6)

holds exactly for all N. Let us denote the application of the N^th partial normal moveout i times by

$${\cal T}_i(t)=\tau_N \tau_N \cdots \tau_N(t)$$ (7)

It holds true that

$${\cal T}_N = T_N.$$ (8)

The superposition of N^th partial normal moveouts may be written

$${\cal T}_i = \sqrt{{\cal T}^2_{i-1}-{x^2 \over Nv^2}}$$ (9)

for $i=1,2,\cdots,N$ and ${\cal T}_0=t$.Let us rewrite this equation into a system of equations:

$$\begin{eqnarray} {\cal T}^2_1&=&{\cal T}^2_0-{x^2 \over Nv^2} \\ {\cal T}^2_2&=&... ...} \\ &\ldots & \\ {\cal T}^2_N&=&{\cal T}^2_{N-1}-{x^2 \over Nv^2}\end{eqnarray}$$ (10) (11) (12) (13)

By summing up all the equations we get  

$${\cal T}^2_N=t^2-N{x^2 \over Nv^2}=t^2-{x^2 \over v^2},$$ (14)

which is the desired result.