Avoid redundant computations in IRR calculation

The IRR computation adds and then subtract the previous iteration value before comparing to the tolerance. We can just compute the delta and compare that to the tolerance instead. This should also make the computation more robust if there is a large difference in magnitude between `x` and `delta`

Author: jlopezpena

numpy_financial/_financial.py

@@ -773,10 +773,10 @@ def irr(values, guess=0.1, tol=1e-12, maxiter=100):
     x = 1 / (1 + guess)
 
     for _ in range(maxiter):
-        x_new = x - (npv_(x) / d_npv(x))
-        if abs(x_new - x) < tol:
-            return 1 / x_new - 1
-        x = x_new
+        delta = npv_(x) / d_npv(x)
+        if abs(delta) < tol:
+            return 1 / (x - delta) - 1
+        x -= delta
 
     return np.nan