I have an interest equation and I need to evaluate the interest increase from certain iterations, using Newton's method.
Eq.:$f(i) = (1-(1-i)^{-10})/i -1400/182.25$
$(i -- plan)$ $(0 -- 0.05)$ $(1 -- 0.05106)$ $(2 -- 0.05107)$ ... But when I apply the equation in Matlab, I can not produce similar values. Could someone tell me how I can develop this?
You have to apply a numerical method. You have $\frac{1-(1\color{red}+i)^{-10}}{i} =1400/182.25 $
First you multiply both sides by $(1+i)^{10}$
$\frac{(1+i)^{10}-1}{i} =1400/182.25\cdot (1+i)^{10}$
Then we use that $i=(1+i)-1$
$\frac{(1+i)^{10}-1}{(1+i)-1} =1400/182.25\cdot (1+i)^{10}$
substituting/simplifying: $1+i=q$ and $C=1400/182.25$
$\frac{q^{10}-1}{q-1} =C\cdot q^{10}$
Multiplying by $q-1$
$q^{10}-1 =C\cdot q^{11}-C\cdot q^{10} \qquad |-q^{10}+1$
$0=C\cdot q^{11}-(C+1)\cdot q^{10}+1 \qquad \color{blue}{(1)}$
Now you have a polynomial with degree 11 with the coefficients $C, C+1$ and $1$.
$f(q)=C\cdot q^{11}-(C+1)\cdot q^{10}+1$
Here you can apply the Newton-Raphson method. You start with
$q_1=q_0-\frac{f(q_0)}{f^{\ '}(q_0)}$, where $x_o$ is the initial value. I assume $i$ is the interest rate. There for $q=1+i$ should not be far away from $1.05$. This would I use as the initial value: $q_0=1.05$.
But you first can also try out some values for $q$ at $ \color{blue}{1}$ for a rough estimation of $q$. If $f(q)$ is close to $0$ then you need not so many iteration.
Hint: $f^{\ '}(q)=11\cdot C\cdot q^{10}-10\cdot (C+1)\cdot q^{9}$