Use Newton’s method to determine the interest rate if the interest was compounded continuously. Consider a bank investment. The initial investment is $10,000. After 25 years, the investment has tripled to$30,000.


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In this problem, the initial invest is given P=10000P=10000. Since after t=25t=25 years A=3P=30000A=3P=30000, and the interest was compounded continuously then

A=Pe1100rt\begin{equation} A=P e^{\frac{1}{100} r t} \end{equation}


e0.25r3=00.25r=ln3r=4ln3\begin{equation} \begin{aligned} & e^{0.25 r}-3=0 \\ \Rightarrow & 0.25 r=\ln 3 \\ \Rightarrow & r=4 \ln 3 \end{aligned} \end{equation}

Now, using the Newton's Iterative Formula

xn+1=xnf(xn)f(xn)=xne0.25xn314e0.25xn\begin{aligned} x_{n+1}&=&x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}\\ &=&x_{n}-\frac{e^{0.25 x_{n}}-3}{\frac{1}{4} e^{0.25 x_{n}}} \end{aligned}

with the initial guess x0=1x_0=1 we have

x1=x0ex0/4314ex0/4=5.27837x2=x1f(x1)f(x1)=4.4853x3=x2f(x2)f(x2)=4.39547x4=x3f(x3)f(x3)=4.39445\begin{equation} \begin{array}{l}{x_{1}=x_{0}-\frac{e^{x_{0} / 4}-3}{\frac{1}{4} e^{x_{0} / 4}}=5.27837} \\ {x_{2}=x_{1}-\frac{f\left(x_{1}\right)}{f^{\prime}\left(x_{1}\right)}=4.4853} \\ {x_{3}=x_{2}-\frac{f\left(x_{2}\right)}{f^{\prime}\left(x_{2}\right)}=4.39547} \\ {x_{4}=x_{3}-\frac{f\left(x_{3}\right)}{f^{\prime}\left(x_{3}\right)}=4.39445}\end{array} \end{equation}

Therefore, the interest is r=4.39445%r=4.39445 \% with more that 4 decimal place accuracy

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