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I'm trying to simplify an expression for reinvesting shares at fixed buy and dividend rates.

Assumptions: $N$ for initial number of shares. $N$ is integer. 2.2p per share is dividend & is fixed. There are $4$ dividends per year, occurring every 3 months. Dividends are reinvested the day of receipt. No charges for reinvesting shares. Buy price per share is fixed at $109.2$p. Broker doesn't allow fractional shares to be bought. Left over money from being unable to buy fractional shares is not reinvested.

First dividend example: $0.022N$p in value. Dividing $0.022N$p by the buy price of $109.2$p gives $\frac{0.022N}{109.2}$ which is the potential amount purchased by the broker (reinvested dividend), but broker doesn't allow fractional shares so the amount of shares bought is actually the floor of this, so: $\lfloor\frac{0.022N}{109.2}\rfloor$. Then the amount of shares owned after the reinvestment is $N+\lfloor\frac{0.022N}{109.2}\rfloor$.

Second dividend example: $0.022(N+\lfloor\frac{0.022N}{109.2}\rfloor)$p in value. Dividing this by the buy price of $109.2$p gives $$\frac{0.022(N+\lfloor\frac{0.022N}{109.2}\rfloor)}{109.2}$$ which is the potential amount purchased by the broker (reinvested dividend), but the amount of shares bought is floor of this, so: $$\lfloor\frac{0.022(N+\lfloor\frac{0.022N}{109.2}\rfloor)}{109.2}\rfloor$$ Then the amount of shares owned after the second reinvestment is $$N+\lfloor\frac{0.022N}{109.2}\rfloor+\lfloor\frac{0.022(N+\lfloor\frac{0.022N}{109.2}\rfloor)}{109.2}\rfloor$$

Then value of those shares owned using the buy price is $$109.2(N+\lfloor\frac{0.022N}{109.2}\rfloor+\lfloor\frac{0.022(N+\lfloor\frac{0.022N}{109.2}\rfloor)}{109.2}\rfloor)p$$

Q1: how can you simplify these two expressions for the shares owed & their value after second dividend reinvested in terms of $N$?

Q2: how can you simplify the formula for the shares owed & their value after $n$ dividends reinvested in terms of $N$?

I have found some inspiration from questions such as Algebraic manipulation of floors and ceilings, however that doesn't explain to me what methods can be used for the simplification.

unseen_rider
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1 Answers1

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Let us define $x_n$ to be the number of shares after the $n$'th dividend and then $x_0=N$. Note that $x_n$ is always an integer, so adding it to another number doesn't affect the fractional part of that other number. That means that $x_{n+1}=x_n+\left\lfloor{\frac{0.022 x_n}{109.2}}\right\rfloor=\left\lfloor x_n + \frac{0.022 x_n}{109.2} \right\rfloor = \left\lfloor \frac{109.222}{109.2} x_n \right\rfloor$. So let us define $r = \frac{109.222}{109.2}=\frac{54611}{54600}$.

In particular we have after the first dividend $x_1=\lfloor r N\rfloor$ shares and after the second dividend $x_2=\lfloor r \lfloor r N \rfloor \rfloor$.

You cannot make the general expression for $x_n$ much nicer other than rewrite it using modulo operations. You can however give bounds. On the one hand, we see that $x_{n+1}\leq r x_n$, so that $x_n\leq r^n N$. On the other hand, we have at most a fractional part of $\frac{54599}{54600}$, so that $x_{n+1}\geq r x_n - \frac{54599}{54600}$. Solving this recurrence gives $x_n\geq r^n \left(N - \frac{54599}{11}\right) + \frac{54499}{11}.$ So to conclude we have $$ r^n \left(N - \frac{54599}{11}\right) + \frac{54599}{11} \leq x_n \leq r^n N. $$

user146125
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