two-sequences-u-n-and-v-n-for-n-N-is-defined-as-u-0-3-u-n-1-1-2-u-n-v-n-and-v-0-4-v-n-1-1-2-u-n-1-v-n-a-calculate-u-1-v-1-u-2-and

Question Number 76630 by Rio Michael last updated on 28/Dec/19

two sequences, (u_{n}) a n d (v_{n}), for n \in N is defined as :

{\begin{cases} u_{0} = 3 \\ u_{n + 1} = \frac{1}{2} (u_{n} + v_{n}) \end{cases} and {\begin{cases} v_{0} = 4 \\ v_{n + 1} = \frac{1}{2} (u_{n + 1} + v_{n}) \end{cases}

a) calculate u_{1}, v_{1}, u_{2} and v_{2}

b) Another sequence (w_{n}), is defined by

w_{n} = v_{n} - u_{n}, \forall n \in N

show that w_{n} is a convegent geometric sequence .

c) Express w_{n} as a function of n and obtain its limits .

d) Study the sense of variation (monotony) of (u_{n}) and (v_{n})

what can you deduce ?

e) Consider another sequence t_{n} defined by

t_{n} = \frac{u_{n} + 2 v_{n}}{3}, \forall n \in N

show that t_{n} is a constant sequence

f) hence obtain the limit of the sequences (u_{n}) and (v_{n})

Answered by mr W last updated on 29/Dec/19

u_{n + 1} = \frac{1}{2} (u_{n} + v_{n}) . . (i)

v_{n + 1} = \frac{1}{2} (u_{n + 1} + v_{n}) \dots (i i)

(i i) - (i) :

v_{n + 1} = \frac{3}{2} u_{n + 1} - \frac{1}{2} u_{n}

\Rightarrow v_{n} = \frac{3}{2} u_{n} - \frac{1}{2} u_{n - 1}

p u t t h i s i n t o (i) :

\Rightarrow 4 u_{n + 1} - 5 u_{n} + u_{n - 1} = 0

4 x^{2} - 5 x + 1 = 0

(4 x - 1) (x - 1) = 0

\Rightarrow x = \frac{1}{4}, x = 1

\Rightarrow u_{n} = \frac{A}{4^{n}} + B

u_{0} = 3 \Rightarrow 3 = A + B

u_{1} = \frac{1}{2} (3 + 4) = \frac{7}{2} \Rightarrow \frac{7}{2} = \frac{A}{4} + B

\Rightarrow - \frac{1}{2} = \frac{3 A}{4} \Rightarrow A = - \frac{2}{3}

\Rightarrow B = 3 + \frac{2}{3} = \frac{11}{3}

\Rightarrow u_{n} = \frac{1}{3} (11 - \frac{2}{4^{n}})

\Rightarrow v_{n} = \frac{1}{3} (11 + \frac{1}{4^{n}})

\dots

(b)

w_{n} = v_{n} - u_{n} = \frac{1}{3} (11 + \frac{1}{4^{n}}) - \frac{1}{3} (11 - \frac{2}{4^{n}})

w_{n} = \frac{1}{4^{n}} \Rightarrow G . P . w i t h c o m m o n r a t i o \frac{1}{4}

\dots .

(e)

t_{n} = \frac{u_{n} + 2 v_{n}}{3} = \frac{\frac{2}{3} (11 + \frac{1}{4^{n}}) + \frac{1}{3} (11 - \frac{2}{4^{n}})}{3}

\Rightarrow t_{n} = \frac{11}{3} = c o n s t a n t

Commented by mr W last updated on 29/Dec/19

a l t e r n a t i v e :

\Rightarrow 4 u_{n + 1} - 5 u_{n} + u_{n - 1} = 0

\Rightarrow 4 u_{n + 1} - 4 u_{n} - (u_{n} - u_{n - 1}) = 0

\Rightarrow (u_{n + 1} - u_{n}) = \frac{1}{4} (u_{n} - u_{n - 1})

\Rightarrow c_{n + 1} = \frac{1}{4} c_{n} \Rightarrow G . P .

\Rightarrow c_{n + 1} = c_{1} {(\frac{1}{4})}^{n}

c_{1} = u_{1} - u_{0} = \frac{7}{2} - 3 = \frac{1}{2}

\Rightarrow c_{n + 1} = \frac{1}{2} {(\frac{1}{4})}^{n}

\Rightarrow u_{n + 1} - u_{n} = \frac{1}{2} {(\frac{1}{4})}^{n}

\Rightarrow \underset{0}{\sum^{n}} u_{n + 1} - \underset{0}{\sum^{n}} u_{n} = \underset{0}{\sum^{n}} \frac{1}{2} {(\frac{1}{4})}^{n}

\Rightarrow u_{n + 1} - u_{0} = \frac{1}{2} \times \frac{1 - {(\frac{1}{4})}^{n + 1}}{1 - \frac{1}{4}} = \frac{2}{3} [1 - {(\frac{1}{4})}^{n + 1}]

\Rightarrow u_{n + 1} = \frac{2}{3} [1 - {(\frac{1}{4})}^{n + 1}] + 3 = \frac{1}{3} (11 - \frac{2}{4^{n + 1}})

o r

\Rightarrow u_{n} = \frac{1}{3} (11 - \frac{2}{4^{n}})

Commented by Rio Michael last updated on 29/Dec/19

thanks sir