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3 years ago

What is the value of a share of Henley Inc. to an investor who requires a 11 percent

Mathematics

1 answer:

Karolina [17]3 years ago

4 0

Answer:

Value of Henley Inc.'s share is $16.76

Step-by-step explanation:

The present value of the dividends over for the three years and the terminal value of the dividends would give us a fair share price that an investor would pay

Year 1 PV of dividends=$1.10/(1+11%)^1=$0.99

Year 2 PV of dividends=$1.10/(1+11%)^2=$0.89

Year 3 PV of dividends=$1.10/(1+11%)^3=$0.80

The terminal value formula=dividend*(1+g)/(r-g)

g is the dividend growth rate of 5%

r is the investor's required rate of return which is 11%

terminal value=$1.10*(1+5%)/(11%-5%)=$19.25

The terminal is discounted to present value using the discount factor of year 3

PV of terminal value =$19.25 /(1+11%)^3=$ 14.08

Total present values=$0.99+$0.89+$0.80+$14.08 =$16.76

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Answer: $-4 \le x \le 5$
Note: to write the domain in interval notation, you'd write [-4,5]
if you need the domain in set-builder notation, then you'd write $\left\{x|x\in\mathbb{R}, \ -4\le x\le 5\right\}$

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Explanation:

The domain is the set of possible x input values. Look at the left most point (-4,-1). The x coordinate here is x = -4. This is the smallest x value allowed. The largest x value allowed is x = 5 for similar reasons, but on the other side of the graph.

So that's how I got $-4 \le x \le 5$ (x is between -4 and 5; inclusive of both endpoints)

Writing [-4,5] for interval notation tells us that we have an interval from -4 to 5 and we include both endpoints. The square brackets mean "include endpoint"

Writing $\left\{x|x\in\mathbb{R}, \ -4\le x\le 5\right\}$ is the set-builder notation way of expressing the domain. The $x\in\mathbb{R}$ portion means "x is a real number"

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$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

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

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

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