# 数学代写|金融数学代写Intro to Mathematics of Finance代考|ACF11003

## 数学代写|金融数学代写Intro to Mathematics of Finance代考|Unknown Time

Problems involving an unknown value of $t$ often lead to equations which can only be solved approximately using numerical methods. The TI BA II Plus, Excel, and MAPLE all have approximation algorithms. Of the three, Excel seems to be the least accurate. While the differences are slight, for large amounts of money, Excel might lead to significant errors.

A common problem concerns replacing a sequence of payments (called an annuity – to be discussed in greater detail later) with a single payment equal to the sum of the numerical values of the other payments. The problem determine the time at which the single payment is to be made in order that the sequence of payments is equal in value to the single payment.

As our first example, suppose that payments in the amounts $p_1, p_2, p_3, \ldots, p_n$ are made at times $t_1, t_2, t_3, \ldots, t_n$. Suppose further that interest is compounded at an effective rate of $i$ per period. We are to find a time, $t$, at which a single payment of $p_1+p_2+p_3+\cdots+p_n=\sum_{k=1}^n p_k$ has the same value as this annuity. We compute our equation of value at the inception of the transaction resulting in the following equation of value. The left side of Equation $3.4$ is the value of the single payment at time $t=0$, while the right side is the value of the sequence of payments at this same time
$$\left(\sum_{k=1}^n p_k\right) v^t=\sum_{k=1}^n p_k v^{t_k}$$
Solving this equation for $t$ yields the following equation:
$$t=\frac{\ln \left(\frac{\sum_{k=1}^n \mathbf{p}k v^{t_k}}{\left(\sum{k=1}^n p_k\right)}\right)}{\ln (v)}$$

## 数学代写|金融数学代写Intro to Mathematics of Finance代考|Doubling Time

$$P_0 a(t)=2 P_0 a\left(t_0\right) P_0(1+i)^t \quad=2 P_0(1+i)^{t_0}(1+i)^t=2(1+i)^{t_0}$$

$$t \ln (1+i)=\ln (2)+t_0 \ln (1+i)\left(t-t_0\right) \quad=\frac{\ln (2)}{\ln (1+i)}$$

$$\Delta t=\frac{\ln (k)}{\ln (1+i)}$$

$$A\left(t_1\right)=k A\left(t_0\right)=\frac{a\left(t_1\right)}{a\left(t_0\right)} A\left(t_0\right) \quad \frac{a\left(t_1\right)}{a\left(t_0\right)}=k=\frac{(1+i)^{t_1}}{(1+i)^{t_0}}=(1+i)^{t_1-t_0} \ln (k)=\left(t_1-t_0\right) \ln (1+i)$$

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