Unit 4

13.10. Suppose that a sequence (ao, a1, a2, ) of real numbers satisfies the recurrence relation an -5an-1+6an-20 for all n> 2

Gwar [14]

a. This recurrence is of order 2.

b. We're looking for a function $A(x)$ such that

$A(x)=\displaystyle\sum_{n=0}^\infty a_nx^n$

Take the recurrence,

$\begin{cases}a_0=a_0\\a_1=a_1\\a_n-5a_{n-1}+6a_{n-2}=0&\text{for }n\ge2\end{cases}$

Multiply both sides by $x^{n-2}$ and sum over all integers $n\ge2$ :

$\displaystyle\sum_{n=2}^\infty a_nx^{n-2}-5\sum_{n=2}^\infty a_{n-1}x^{n-2}+6\sum_{n=2}^\infty a_{n-2}x^{n-2}=0$

Pull out powers of $x$ so that each summand takes the form $a_kx^k$ :

$\displaystyle\frac1{x^2}\sum_{n=2}^\infty a_nx^n-\frac5x\sum_{n=2}^\infty a_{n-1}x^{n-1}+6\sum_{n=2}^\infty a_{n-2}x^{n-2}=0$

Now shift the indices and add/subtract terms as needed to get everything in terms of $A(x)$ :

$\displaystyle\frac1{x^2}\left(\sum_{n=0}^\infty a_nx^n-a_0-a_1x\right)-\frac5x\left(\sum_{n=0}^\infty a_nx^n-a_0\right)+6\sum_{n=0}^\infty a_nx^n=0$

$\displaystyle\frac{A(x)-a_0-a_1x}{x^2}-\frac{5(A(x)-a_0)}x+6A(x)=0$

Solve for $A(x)$ :

$A(x)=\dfrac{a_0+(a_1-5a_0)x}{1-5x+6x^2}\implies\boxed{A(x)=\dfrac{a_0+(a_1-5a_0)x}{(1-3x)(1-2x)}}$

c. Splitting $A(x)$ into partial fractions gives

$A(x)=\dfrac{2a_0-a_1}{1-3x}+\dfrac{3a_0-a_1}{1-2x}$

Recall that for $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n$

so that for $|3x|$ and $|2x|$ , or simply $|x|$ , we have

$A(x)=\displaystyle\sum_{n=0}^\infty\bigg((2a_0-a_1)3^n+(3a_0-a_1)2^n\bigg)x^n$

which means the solution to the recurrence is

$\boxed{a_n=(2a_0-a_1)3^n+(3a_0-a_1)2^n}$

d. I guess you mean $a_0=2$ and $a_1=5$ , in which case

$\boxed{\begin{cases}a_0=2\\a_1=5\\a_2=13\\a_3=35\\a_4=97\\a_5=275\end{cases}}$

e. We already know the general solution in terms of $a_0$ and $a_1$ , so just plug them in:

$\boxed{a_n=2^n+3^n}$

8 0

3 years ago

Find the 6th term of the geometric sequence whose common ratio is 3/2 and whose first term is 5

Fynjy0 [20]

Answer:

a(6) = 37.97

Step-by-step explanation:

If the first term a(1) is 5 and the common ratio r is 3/2, then the general formula for this geometric sequence is

a(n) = 5*r^(n - 1).

Thus, the 6th term is a(6) = 5*(3/2)^(6 - 1), or a(6) = 5(3/2)^5 = 37.97

5 0

3 years ago

What is the prime factorization of 90? 2×2×3×5 2×5×9 2×3×3×5 2×3×5

Firdavs [7]

It’s 2x3x3x5 because 9 isn’t prime

3 0

4 years ago

Read 2 more answers

1. Solve forx:<br> 5(x - 8) + 7 = 2

ch4aika [34]

Answer: x=7

Step-by-step explanation:

(5x-40)+7 = 2

5x-33 = 2

5x = 35

x = 35/5

x = 7

8 0

3 years ago

Read 2 more answers

An astronaut on the moon throws a baseball upward. The astronaut is 6 ft 6 in tall, and the initial velocity of the ball is 40 f

weqwewe [10]

Answer:

t = 0.293 s and 14.52 s

Step-by-step explanation:

The heights of the ball in feet is given by the equation as follows :

$S = -2.7t^2+40t+6.5$

Where t is the number of seconds after the ball was thrown.

We need to find is the ball 18 ft above the moon's surface.

Put S = 18 ft

$-2.7t^2+40t+6.5=18\\\\-2.7t^2+40t=18-6.5\\\\-2.7t^2+40t-11.5=0$

It is a quadratic equation. Its solution is given by :

$t=\dfrac{-40+\sqrt{40^{2\ }-4\left(-2.7\right)\left(-11.5\right)}}{2\left(-2.7\right)},\dfrac{-40-\sqrt{40^{2\ }-4\left(-2.7\right)\left(-11.5\right)}}{2\left(-2.7\right)}\\\\t=0.293\ s,14.52\ s$

So, the ball is at first 0.293 s and then 14.52 s above the Moon's surface.

8 0

3 years ago