Update appendix-b.md

Author: pavly-gerges

discrete-maths/appendix-b.md

@@ -64,8 +64,9 @@ $$Since, f(N) = C(N) * \sum_{n=1}^N N_c$$
 - Back-substitution yields: $$N_c = N^2 * (N_{c_b} + N_{\phi}) = N^2 * (N_c + N_{\phi}) = N^2 * (N * (N_{c_c} + N''\_{\phi}) + N'\_{\phi})$$
 $$= N^3 * (N_{c_c} + N''\_{\phi}) + N^2 * N'\_{\phi}$$
 - The following formulas are derivable:
-$$ N_{c_c} = [(N_c - N^2 * N'\_{\phi}) / N^3] - N''\_{\phi} $$
-$$ N''\_{\phi} = [(N_c - N^2 * N'\_{\phi}) / N^3] - N_{c_c} $$
+$$1)\ N_{c_c} = [(N_c - N^2 * N'\_{\phi}) / N^3] - N''\_{\phi}$$
+$$2)\ N''\_{\phi} = [(N_c - N^2 * N'\_{\phi}) / N^3] - N_{c_c}$$
+$$3)\ N'\_{\phi}= [N_c - N^3 * (N_{c_c} + N''\_{\phi})] / N^2$$
 
 - Now, if `command()` has a clock-complexity of (1):
 $$N_c = N^3 * (N_{c_c} + N''\_{\phi}) + N^2 * N'\_{\phi} = N^3 * (1 + 0) + N^2 * 0 = N^3$$