Question

47. A new employee at an exciting new software company starts with a salary of550,000and is promised that at the end of each year her salary will be double her salary of the previous year, with an extra increment of$ 10,000 for each year she has been with the company.

a) Construct a recurrence relation for her salary for hern th year of employment.

b) Solve this recurrence relation to find her salary for hern th year of employment.

Some linear recurrence relations that do not have constant coefficients can be systematically solved. This is the case for recurrence relations of the form \(f(n){a_n} = g(n){a_{n - 1}} + h(n)\)Exercises 48-50 illustrate this.

Short answer

(a) $a_n=2a_{n-1}+10000(n-1),a_1=50000$

(b) $a_n=35000·2^n-10000n-10000=70000·2^{n-1}-10000n-10000$

Step-by-step solution

Linear Homogeneous Recurrence Relations with Constant Coefficients definition

Let\({c_1}\)and\({c_2}\)be real numbers. Suppose that\({r^2} - {c_1}r - {c_2} = 0\)has two distinct roots\({r_1}\)and\({r_2}\). Then the sequence\(\left\{ {{a_n}} \right\}\)is a solution of the recurrence relation\({a_n} = {c_1}{a_{n - 1}} + {c_2}{a_{n - 2}}\)if and only if\({a_n} = {\alpha _1}r_1^n + {\alpha _2}r_2^n\)for\(n = 0,1,2, \ldots \), where\({\alpha _1}\)and\({\alpha _2}\)are constants.

Use Recurrence Relations for part (a)           

(a) Derivation recurrence relation

Let $a_n$represent the salary at the n th year.

First case The salary doubles in comparison to the previous year. The salary of the previous year is $a_{n-1}$.

$2a_{n-1}$

Second case An extra 10000 dollars is added to the salary for each of the $n-1$ years that she has been with the company.

$10000(n-1)$

Add all number of pairs for each case:

$a_n=2a_{n-1}+10000(n-1)$

Initial conditions

At the beginning of the first year, the salary is$50,000a_1=50000$

Use Recurrence Relations for part (b)

(b) Let \({a_n} = r\) and \({a_{n - 1}} = 1\) (other functions of \(n\) are considered to be 0 ) and \(r = 2\)

The solution of the recurrence relation is then of the form\({a_n} = {\alpha _1}r_1^n + {\alpha _2}nr_1^n + \ldots + {\alpha _k}{n^{k - 1}}r_1^n\) with\({r_1}\) a root with multiplicity\(k\) of the characteristic equation.

\(a_n^{(h)} = \alpha \cdot {2^n}\)

Thus the solution of the homogeneous recurrence relation is\(a_n^{(h)} = \alpha \cdot {2^n}\)

\(F(n) = 10000(n - 1) = 10000(n - 1) \cdot {1^n}\)

If\(F(n) = \left( {{b_t}{n^t} + {b_{t - 1}}{n^{t - 1}} + \ldots + {b_1}n + {b_0}} \right){s^n}\) and\(s\) is not a root of the characteristic equation, then\(\left( {{p_t}{n^t} + {p_{t - 1}}{n^{t - 1}} + \ldots + {p_1}n + {p_0}} \right){s^n}\) is the particular solution.

If\(F(n) = \left( {{b_t}{n^t} + {b_{t - 1}}{n^{t - 1}} + \ldots + {b_1}n + {b_0}} \right){s^n}\) and\(s\) is\({\bf{a}}\)root of the characteristic equation with multiplicity\({m_r}\) then\({n^m}\left( {{p_t}{n^t} + {p_{t - 1}}{n^{t - 1}} + \ldots + {p_1}n + {p_0}} \right){s^n}\)is the particular solution.

1 is not a root of the characteristic equation:

\(a_n^{(p)} = \left( {{p_1}n + {p_0}} \right) \cdot {1^n} = {p_1}n + {p_0}\)

The particular solution needs to satisfy the recurrence relation:

\(\begin{array}{l}{a_n} = 2{a_{n - 1}} + 10000(n - 1){p_1}n + {p_0}\\{a_n} = 2\left( {{p_1}(n - 1) + {p_0}} \right) + 10000(n - 1){p_1}n + {p_0}\\{a_n} = 2{p_1}n - 2{p_1} + 2{p_0} + 10000n - 100000\\{a_n} = \left( {{p_1} + 10000} \right)n + \left( { - 2{p_1} + {p_0} - 10000} \right){\rm{ }}\end{array}\)

All coefficients then need to be zero\({p_1} + 10000 = 0 - 2{p_1} + {p_0} - 10000 = 0{\rm{ }}\)

Solve each equation:

\(\begin{array}{l}{\rm{ }}{p_1} = - 10000\\{p_0} = 10000 + 2{p_1}\\{p_0} = 10000 + 2( - 10000)\\{p_0} = - 10000\end{array}\)

The particular solution then becomes:

\(a_n^{(p)} = {p_1}n + {p_0} = - 10000n - 10000\).

Solve the relation

The solution of the nonhomogeneous linear recurrence relation is the sum of the homogeneous\(a_n^{(h)}\) and particular solution\(a_n^{(p)}\)

\({a_n} = a_n^{(h)} + a_n^{(p)} = \alpha \cdot {2^n} - 10000n - 10000\)

Evaluate\({a_n} = \alpha \cdot {2^n} - 10000n - 10000\) at\(n = 1\)

Equation is:\(50000 = {a_1} = 2\alpha - 10000 - 10000\)

Combine like terms:\(50000 = 2\alpha - 20000\)

Add 20000 to each side\(70000 = 2\alpha \)

Divide each side by\(2\):\(35000 = \alpha \)

The solution of the nonhomogeneous linear recurrence relation then becomes:

\(\begin{array}{l}{a_n} = \alpha \cdot {2^n} - 10000n - 10000\\{a_n} = 35000 \cdot {2^n} - 10000n - 10000\\{a_n} = 70000 \cdot {2^{n - 1}} - 10000n - 10000\end{array}\)