#X_(n+1)-aX_n+2=0#
Which are the set values of "a" for the string "Xn" is descending?

Question

#X_(n+1)-aX_n+2=0#

Luke Alvarez · Accepted Answer

See below.

#X_(n+1)-aX_n+2=0# Is a linear non homogeneous difference equation. It's solution can be composed of a solution for the homogeneous #X_n^h# such that #X_(n+1)^h-aX_n^h=0# plus a particular solution #X_n^p# for the non homogeneous equation such that #X_(n+1)^p-aX_n^p+2=0# For the homogeneous solution the proposal is #X_n^h=Ca^n#. Substituting #Ca^(n+1)-aCa^n= Ca^(n+1)-C a^(n+1)=0# Now for the particular we propose #X_n^p=C_na^n# then #C_(n+1)a^(n+1)-aC_na^n=a^(n+1)(C_(n+1)-C_n)=2# so #C_(n+1)-C_n=2/a^(n+1)# then beginning with #C_0# we have #C_1= C_0 + 2/a# #C_2=C_1+2/a^2 = C_0 +2/a+2/a^2# #...# and then #C_n = C_0+2 sum_(k=1)^na^(-k)# and finally #X_n = X_n^h+X_n^p=(C_0+2 sum_(k=1)^na^(-k))a^n=C_0a^n+2sum_(k=1)^na^k# If we need that #X_n < X_(n-1)# then #C_0a^n+2sum_(k=1)^na^k < C_0a^(n-1)+2sum_(k=1)^(n-1)a^k# or #C_0a^n+2a^n < C_0a^(n-1)# or #a < C_0/(C_0+2)# Here #C_0# is the initial condition.

Charles Arocha · Answer

undefined The set of values for $ a $ such that the sequence $ X_n $ is descending depends on the initial conditions and the specific characteristics of the sequence. Without further information about the initial values of the sequence $ X_n $, it is not possible to determine a specific set of values for $ a $ that guarantees a descending sequence.

#X_(n+1)-aX_n+2=0# Which are the set values of "a" for the string "Xn" is descending?

#X_(n+1)-aX_n+2=0#