Solve #f(x)=f(x-2)+1 # and #f(3)=5 # ?

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Answer 1 · 2017-12-11T12:42:34

See below.

Considering

#f(x)=f(x-2)+1 # and #f(3)=5 #

as a difference equation we have

#f_n = f_(n-2) + 1#

This recurrence equation is a linear one so the answer can be composed as

#f_n = f_n^h + f_n^p# with

#f_n^h =f_(n-2)^h# and

#f_n^p=f_(n-2)^p +1#

Considering #f_n^h = C alpha^n# we have

#Calpha^n = C alpha^(n-2) rArr C alpha^n(1-alpha^-2) rArr alpha = pm 1#

and

#f_n^p = (2n-1)/4# and then

#f_n = C_1 + C_2(-1)^n +(2n-1)/4#

we have also

#f_3=C_1 + C_2(-1)^3 +(2xx3-1)/4 = C_1-C_2 + 5/4 = 5# and then

#f_4 = 5-5/4+2C_2+7/4 = 11/2+2C_2#