A Removable Singularity in LeadLag Coefficients 

As discussed in section 2.2.4 of LeadLag Algorithms, a recursive simulation of a firstorder leadlag transfer function 

_{} 

can be written in the form 

_{} 

where the subscripts "c" and "p" denote current and past values respectively. If the time constants t_{D} and t_{N} are variable, and vary linearly over each time increment Dt, then the coefficients of the recurrence can be written as 

_{} 

where 
_{} 

On the surface it may appear that B is infinite when _{}, but we can show that this is due to a removable singularity, and B is actually wellbehaved at _{}. To show this, we first rewrite A in the form 

_{} 

Assuming _{}, we can expand the binomial to write this as 

_{} 

Extending the series and simplifying, we have 

_{} 

We now rewrite B in the form 

_{} 

Notice that as _{} goes to 1, the coefficient A goes to Dt/t_{DP}, so the quantity in the right hand parentheses goes to zero. In the same condition the quantity in the left hand parentheses goes to infinity. Our assertion is that the product of these two quantities approaches a finite value in this limit. To determine this limiting value, we can substitute the series expression for A into the above expression for B to give 

_{} 

We can therefore cancel a factor of _{} from the numerator and denominator, leaving us with 

_{} 

Now, setting _{} to 1 (which makes A equal to Dt/t_{DP}), we have 

_{} 

Thus when _{} to 1 the coefficient B has the value given by 

_{} 

Note that the original restriction _{} again corresponds to the condition of convergence, since _{} equals 1, so the absolute value of Dt is less than t_{DP}. To evaluate the infinite sum, let x denote the ratio Dt/t_{DP}, and differentiate the summation as follows. 

_{} 

The summation in the right hand expression is the series expansion of ln(1x), so we have 

_{} 

Making use of the wellknown integral 

_{} 

with a = b = 1, we have 

_{} 

and therefore setting z = 1  x (which gives dz = dx) we arrive at 

_{} 

Combining this with the 1/x term in the derivative of our infinite sum, and simplifying, we have the overall integral 

_{} 

where the constant of integration is determined by the fact that our original summation vanishes as x = 0, and noting that the product (1/x  1)ln(1  x) goes to 1 as x goes to zero. Thus our summation can be written in closed form as 

_{} 

Therefore, in terms of our original problem, the singularity in B when _{} equals 1 can be removed by assigning B the value 

_{} 

which was to be shown. 
