Question #147114
Solve the following recurrence relations
i) Fn= Fn-1 +Fn-2 where a1=a2=1
ii) an=2an-1 - an-2 +2 where a1 = 1 and a2 = 5
1
Expert's answer
2020-11-30T10:37:25-0500

i) Write down the characteristic equation:

x2=x+1x^2 = x + 1

x2x1=0x^2 -x -1 =0

x1=1+52=ϕ; x2=152=ψx_1 =\frac{1 + \sqrt5}2 = \phi ; \ x_2 = \frac {1 -\sqrt 5}2 =\psi

Fn=ax1n+bx2nF_n = ax_1^n +bx_2^n

aϕ+bψ=1 and  aϕ2+bψ2=1a*\phi +b*\psi = 1\ and\ \ a*\phi^2 +b*\psi^2 = 1

aϕ=1bψ;a*\phi = 1 - b*\psi;

(1bψ)ϕ+bψ2=1(1-b*\psi)*\phi +b*\psi^2 = 1

ϕψ=154=1;\phi*\psi = \frac{1- 5}4 = -1;

ϕ+b+bψ2=1\phi +b + b*\psi^2 = 1

b(1+ψ2)=1ϕb*(1+\psi^2) = 1 -\phi

ψ2>0;  1+ψ2>0\psi^2 > 0;\ \ 1+ \psi^2 >0

b=1ϕ1+ψ2=11+521+(152)2=b = \frac {1-\phi}{1+\psi^2} = \frac {1- \frac{1+\sqrt 5}2}{1+ (\frac{1-\sqrt5}2)^2} =

=15225254=1525(15)2=15= \frac{\frac{1-\sqrt5}{2}}{\frac{2*5 -2*\sqrt5}4} = \frac{\frac{1-\sqrt5}{2}}{-\sqrt5*\frac{(1 - \sqrt5)}2}=-\frac {1}{\sqrt5}

ϕ>0;  a=1+ψ5ϕ=1+15251+52=\phi > 0;\ \ a = \frac{1 + \frac{\psi}{\sqrt5}}{\phi} =\frac{1+\frac{\frac{1-\sqrt5}{2}}{\sqrt5}}{\frac{1+\sqrt5}{2}}=

=25+15251+52=1+5251+52==\frac{\frac{2*\sqrt5+1-\sqrt5}{2\sqrt5}}{\frac{1+\sqrt5}{2}} =\frac{\frac{1+\sqrt5}{2\sqrt5}}{\frac{1+\sqrt5}{2}}=

=151+521+52=15=\frac{1}{\sqrt5}*\frac{\frac{1+\sqrt5}{2}}{\frac{1+\sqrt5}{2}} = \frac{1}{\sqrt5}

Fn=ϕn5ψn5=ϕnψn5;F_n = \frac{\phi^n}{\sqrt5} - \frac{\psi^n}{\sqrt5}= \frac{\phi^n - \psi^n}{\sqrt5};


ii) a3=251+2=10+1=11a_3 = 2*5 -1 + 2 = 10 + 1 = 11

an+1=2anan1+2;a_{n+1} = 2a_{n} - a_{n-1} + 2;

an=2an1an2+2;a_n = 2a_{n-1} - a_{n-2} + 2;

From the first equation subtract the second:

an+1an=2anan1+22an1+an22a_{n+1} - a_n = 2a_n - a_{n-1} + 2 - 2a_{n-1} + a_{n-2} - 2

an+1=3an3an1+an2 for n3a_{n+1} = 3a_n - 3a_{n-1} + a_{n-2} \ for \ n \ge 3

Therefore, for n4:an=3an13an2+an3;n \ge 4: a_n = 3a_{n-1} - 3a_{n-2} + a_{n-3};

Write down the characteristic equation:

x33x2+3x1=0x^3 - 3x^2 +3x -1 = 0

(x1)3=0(x-1)^3= 0

x1,2,3=1;x_{1,2,3}= 1;

an=k11n+k2n1n+k3n21n=a_n = k_1*1^n + k_2*n*1^n + k_3*n^2*1^n =

=k1+k2n+k3n2;=k_1+ k_2*n + k_3*n^2;


The system of three equations:

k1+k2+k3=a1=1k_1+ k_2 + k_3= a_1 = 1

k1+k22+k34=a2=5k_1+ k_2*2 + k_3*4= a_2 = 5

k1+k23+k39=a3=11k_1+ k_2*3 + k_3*9= a_3 = 11


Subtract the first from the second:

k2+k33=4k_2 + k_3*3= 4 (iv)


Subtract the second from the third:

k2+k35=6k_2 + k_3*5= 6 (v)


Subtract (iv) from (v):

k32=2k_3*2 = 2

k3=1k_3 = 1


From this and (iv):

k2+3=4k_2 +3 = 4

k2=1k_2 = 1


From these and the first:

k1+1+1=1k_1 + 1 +1 = 1

k1=1k_1=-1


an=1+n+n2=n2+n1.a_n = -1 + n + n^2 = n^2 +n -1.


Answer: i) Fn=ϕnψn5F_n = \frac{\phi^n - \psi^n}{\sqrt5}

ii) an=n2+n1a_n = n^2 +n -1



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