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hello
please check my answer and let me know if there are any errors and if there are better ways to solve ......
Solve the recurrence relation
an = an−1 + 20an−2
with a0 = −1 and a1 = 13.
Step – I : Find Characteristic equation and solve for its roots:
an = an−1 + 20an−2 ------equation (0)
First of all make corresponding characteristic equation of equation 0 and find roots of that equation
let
an= r2
an-1 =r
an-2 =r0
Substituting above values in equation (0)
r2 =r + 20
r2-r-20=0 is the characteristic equation.
r2-5r+4r-20=0
r(r-5)+4(r-5)=0
(r-5 ) (r+4)=0
r1 = 5 , r2 = -4 are the roots of this characteristic equation
Step– II : Find Constants’ value using Initial Condition
So solution of this polynomial would be
an = c1 r1n + c2 r2n ------------------equation (1)
Where c1 & c2 are some arbitrary constants .
Now we have to find values of c1 and c2 using Initial Condition.
As a0 = −1 and a1 = 13.
So put n=0 in equation (1)
a0= c1 r10 + c2 r20
a0= c1 + c2
-1 = c1 + c2 ---------equation (2)
Now put n=1 in equation (1)
a1 = c1 r11 + c2 r21
13= c1 (5)1 + c2 (-4)1 --------------equation (3)
Multiply equation (2) by 4
-4= 4c1 + 4c2 -------------------------------------------------equation (4)
Add equation (3) and (4) to get the values of c1 & c2.
-4= 4c1 + 4c2
13= 5c1 - 4c2
9 = 9c1
c1=1
put c1=1 in equation (2)
c2=-2
Step-III ..Substitute values of c1 , c2 , r1 and r2 to get the final solution
so substituting these values in equation
an = 1.5n + (-2) 4n
an = 5n – (2).4n is the final solution of this recurrence relation
please check my answer and let me know if there are any errors and if there are better ways to solve ......
Solve the recurrence relation
an = an−1 + 20an−2
with a0 = −1 and a1 = 13.
Step – I : Find Characteristic equation and solve for its roots:
an = an−1 + 20an−2 ------equation (0)
First of all make corresponding characteristic equation of equation 0 and find roots of that equation
let
an= r2
an-1 =r
an-2 =r0
Substituting above values in equation (0)
r2 =r + 20
r2-r-20=0 is the characteristic equation.
r2-5r+4r-20=0
r(r-5)+4(r-5)=0
(r-5 ) (r+4)=0
r1 = 5 , r2 = -4 are the roots of this characteristic equation
Step– II : Find Constants’ value using Initial Condition
So solution of this polynomial would be
an = c1 r1n + c2 r2n ------------------equation (1)
Where c1 & c2 are some arbitrary constants .
Now we have to find values of c1 and c2 using Initial Condition.
As a0 = −1 and a1 = 13.
So put n=0 in equation (1)
a0= c1 r10 + c2 r20
a0= c1 + c2
-1 = c1 + c2 ---------equation (2)
Now put n=1 in equation (1)
a1 = c1 r11 + c2 r21
13= c1 (5)1 + c2 (-4)1 --------------equation (3)
Multiply equation (2) by 4
-4= 4c1 + 4c2 -------------------------------------------------equation (4)
Add equation (3) and (4) to get the values of c1 & c2.
-4= 4c1 + 4c2
13= 5c1 - 4c2
9 = 9c1
c1=1
put c1=1 in equation (2)
c2=-2
Step-III ..Substitute values of c1 , c2 , r1 and r2 to get the final solution
so substituting these values in equation
an = 1.5n + (-2) 4n
an = 5n – (2).4n is the final solution of this recurrence relation
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